Puzzles- contd

[pavalamani pragasam]

Einstein's Riddle

ALBERT EINSTEIN'S RIDDLE

ARE YOU IN THE TOP 2% OF INTELLIGENT PEOPLE IN THE WORLD?

SOLVE THE RIDDLE AND FIND OUT.


There are no tricks, just pure logic, so good luck and don't give up.


1. In a street there are five houses, painted five different colours.

2. In each house lives a person of different nationality

3. These five homeowners each drink a different kind of beverage, smoke different brand of cigar and keep a different pet.


THE QUESTION: WHO OWNS THE FISH?


HINTS


1. The British man lives in a red house.
2. The Swedish man keeps dogs as pets.
3. The Danish man drinks tea.
4. The Green house is next to, and on the left of the White house.
5. The owner of the Green house drinks coffee.
6. The person who smokes Pall Mall rears birds.
7. The owner of the Yellow house smokes Dunhill.
8. The man living in the center house drinks milk.
9. The Norwegian lives in the first house.
10. The man who smokes Blends lives next to the one who keeps cats.
11. The man who keeps horses lives next to the man who smokes Dunhill.
12. The man who smokes Blue Master drinks beer.
13. The German smokes Prince.
14. The Norwegian lives next to the blue house.
15. The Blends smoker lives next to the one who drinks water.

ALBERT EINSTEIN WROTE THIS RIDDLE EARLY DURING THE 19th CENTURY.

HE SAID THAT 98% OF THE WORLD POPULATION WOULD NOT BE ABLE TO SOLVE IT.


(Happy to inform our family members have found place in that prestigious 2%! :D )

[PatchyBoy]

That one is a classic.

Here is one from me:

John and Paul live in a village. One day John decides to walk to the nearby town, but Paul wants to stay back.

So, John starts walking towards the town at 10 KMPH. After 1 hour, Paul feels lonely and decides to also go. So, he starts walking towards the town at 15 KMPH. At the sametime, Paul's dog starts running ahead of him at 20 KMPH.

The moment the dog catches up with John, it turns around and starts running back to Paul, still at 20 KMPH. The dog continues the running between the two men till Paul catches up with John.

What is the total distance travelled by the Dog?

Rajan

[pavalamani pragasam]

Usually I'm averse to using my grey cells in such mathematical calculations. Plain lazy! Incorrigibly so! The Einstein puzzle too I was not inclined to try at first and watched happily as our children finished solving it one by one. But when my younger sister, a doctor, too found time to solve it I had no option than to prove my mettle! I never budge out of my sloth unless properly challenged! :noteeth:

Hope al the stalwart hubbers who made this thread abuzz with brilliant activity come back!

[pavalamani pragasam]

Two from my grandsons:
1.Why do we feel tired on April 1st?
2.What is the difference between Prince Charles and a ball?

[rajraj]

That one is a classic.

Here is one from me:

John and Paul live in a village. One day John decides to walk to the nearby town, but Paul wants to stay back.

So, John starts walking towards the town at 10 KMPH. After 1 hour, Paul feels lonely and decides to also go. So, he starts walking towards the town at 15 KMPH. At the sametime, Paul's dog starts running ahead of him at 20 KMPH.

The moment the dog catches up with John, it turns around and starts running back to Paul, still at 20 KMPH. The dog continues the running between the two men till Paul catches up with John.

What is the total distance travelled by the Dog?

Rajan

Going back to high school days ! :lol:

Assume x to be the time it takes Paul to catch up with John.

Paul walks 15x miles to catch up with John and during that time John walks 10+10x miles.

10+10x = 15x or x= 2 hours

In two hours the dog runs 40 KM

Of course, high school students would have complained that the question paper was very difficult ! :lol:

[rajraj]

Here is one for you:

What is special about the number 81, apart from being the square of 9 ?

:)

[GP]

Einstein's Riddle

ALBERT EINSTEIN'S RIDDLE

ARE YOU IN THE TOP 2% OF INTELLIGENT PEOPLE IN THE WORLD?

SOLVE THE RIDDLE AND FIND OUT.


There are no tricks, just pure logic, so good luck and don't give up.


1. In a street there are five houses, painted five different colours.

2. In each house lives a person of different nationality

3. These five homeowners each drink a different kind of beverage, smoke different brand of cigar and keep a different pet.


THE QUESTION: WHO OWNS THE FISH?


HINTS


1. The British man lives in a red house.
2. The Swedish man keeps dogs as pets.
3. The Danish man drinks tea.
4. The Green house is next to, and on the left of the White house.
5. The owner of the Green house drinks coffee.
6. The person who smokes Pall Mall rears birds.
7. The owner of the Yellow house smokes Dunhill.
8. The man living in the center house drinks milk.
9. The Norwegian lives in the first house.
10. The man who smokes Blends lives next to the one who keeps cats.
11. The man who keeps horses lives next to the man who smokes Dunhill.
12. The man who smokes Blue Master drinks beer.
13. The German smokes Prince.
14. The Norwegian lives next to the blue house.
15. The Blends smoker lives next to the one who drinks water.

ALBERT EINSTEIN WROTE THIS RIDDLE EARLY DURING THE 19th CENTURY.

HE SAID THAT 98% OF THE WORLD POPULATION WOULD NOT BE ABLE TO SOLVE IT.


(Happy to inform our family members have found place in that prestigious 2%! :D )
German.

Yellow House: Norwegian drinks water, smokes Dunhills, and keeps cats.
Blue House: Danish drinks tea, smokes Blends, and keeps horses.
Red house: Englishman drinks milk, smokes PallMalls, and keeps birds.
White house:Swede drinks bier, smokes BlueMasters, and keeps dogs.
Green house: German drinks coffee, smokes Princes, and keeps fish.

[Benny Lava]

That one is a classic.

Here is one from me:

John and Paul live in a village. One day John decides to walk to the nearby town, but Paul wants to stay back.

So, John starts walking towards the town at 10 KMPH. After 1 hour, Paul feels lonely and decides to also go. So, he starts walking towards the town at 15 KMPH. At the sametime, Paul's dog starts running ahead of him at 20 KMPH.

The moment the dog catches up with John, it turns around and starts running back to Paul, still at 20 KMPH. The dog continues the running between the two men till Paul catches up with John.

What is the total distance travelled by the Dog?

Rajan

In one hours time the distance between John and Paul is 10 Kms. John approaches with a relative speed of 5 Kmph so it would take him 2 hours to reach John.

In 2 hours time, assuming that the dog wouldn't have to decelerate while turning back and changing directions, it would have covered 20 kmph x 2 hours = 40 Kms.

Of couse Rajraj solved it elegantly already but I did not use the dreaded alphabet 'x' so I thought I'wd post it :lol:

[Benny Lava]

Here is one easy questioon...

A craftsman has 5 short pieces of chains as shown below. Each piece contains 3 rings. He has to connect them all into one long chain.

OOO OOO OOO OOO OOO

Every operation that he performs on the rings is going to cost him lots of time (Note that unlinking a ring is one operation and linking it to another ring is another operation). What is the minimum number of operations needed to link the chain together and how?

For sake of explaining, you can number the rings 1 to 15 and elaborate.

[Benny Lava]

Here is one for you:

What is special about the number 81, apart from being the square of 9 ?

:)

Rajraj, seems like your question cannot be answered without Google or Wiki's help... Any clues please? Or else you will get all sorts of imaginative answers :P

[GP]

Unlink 4 & 6, link it to 3 & 7,
Unlink 10 & 12, link it to 9 & 13

8 operations

:?

[GP]

Here is one for you:

What is special about the number 81, apart from being the square of 9 ?

:)

Rajraj, seems like your question cannot be answered without Google or Wiki's help... Any clues please? Or else you will get all sorts of imaginative answers :P
Rajraj, seems like your question cannot be answered with Google or Wiki's help... Any clues please? Or else you will get all sorts of imaginative answers :P

[Benny Lava]

Still it can be reduced GP.. our craftsman is a real miser! :P

[rajraj]

OOO OOO OOO OOO OOO




ooo ooo ooo ooo c c c


ooo ooo ooo ooo
---c---c---c

ooooooooooooooo



6 ! :)

[Benny Lava]

OOO OOO OOO OOO OOO




ooo ooo ooo ooo o o o


ooo ooo ooo ooo
---o---o ---o

ooooooooooooooo


6 ! :)

Again, you presented the solution very elegantly Rajraj sir.. :thumbsup:

P.S: Quote panradhukula count onnu yeriduchu? :P

[rajraj]

Here is one for you:

What is special about the number 81, apart from being the square of 9 ?

:)

Rajraj, seems like your question cannot be answered without Google or Wiki's help... Any clues please? Or else you will get all sorts of imaginative answers :P

Try to express 81, a square, as sum of squares and try to express a few other squares as sum of squares ! :)

[rajraj]

Again, you presented the solution very elegantly Rajraj sir.. :thumbsup:


Thanks ! Retire aanaal kooda vaathiyaar velai ennai vittu pogalai ! :lol:

[Benny Lava]

Here is one for you:

What is special about the number 81, apart from being the square of 9 ?

:)

Rajraj, seems like your question cannot be answered without Google or Wiki's help... Any clues please? Or else you will get all sorts of imaginative answers :P

Try to express 81, a square, as sum of squares and try to express a few other squares as sum of squares ! :)

Rajraj, This is what I could randomly identify in the proximity of 9:

13^2 = 12^2 + 5^2
11^2 = 9^2 + 6^2 + 2^2
10^2 = 6^2 + 4^2
9^2 = 6^2 + 5^2 + 4^2 + 2^2
7^2 = 2^2 + 3^2 + 6^2
5^2 = 3^2 + 4^2

I am ignoring repeat square numbers (like 3^2 = 1^2 + 2^2 + 2^2) otherwise I think we can express a lot of other numbers as square of other numbers. But I am not able to see any pattern here :oops:

[NM]

My brain can only see this : :lol:
81 = the number whereby it is the square of sum of its digits? = (1+8)^2? :oops: can't be that simple, right? :lol:

[rajraj]

My brain can only see this : :lol:
81 = the number whereby it is the square of sum of its digits? = (1+8)^2? :oops: can't be that simple, right? :lol:

ammaa sonnapadi uLundhu saappittirukkaNum ! :lol:


It is the only square of a square (4th power) that can be expressed as a sum of squares ! :) I think that was what the book I browsed said! :)

3^4 = 8^2 + 4^2 + 1^2


If I run across the book again I will verify ! :)

May be somebody can write a script (program) to verify the assertion! :)

Benny Lava : Good job! You get an A ! :lol:

[NM]

:lol: :lol: raj-uncle :lol: unga solution en solution-Oda simple :lol: :lol: athaan en aRivu-kku etta villai :lol:

[rajraj]

:lol: :lol: raj-uncle :lol: unga solution en solution-Oda simple :lol: :lol: athaan en aRivu-kku etta villai :lol:

You were on the right track! :)

81 = (1+8)^2 = (1^2 + 2x1x8 +8^2) = 1^2 + 16 + 8^2 ) =

1^2 + 4^2 + 8^2

:)

[NM]

uncle - athellaam intha notebook-ezhuthi vechchen but thought that's silly :oops: :lol: thanks Uncle :)
pEthi vanthaal, nanna solli kodunggO! :lol:

[GP]

Simple one.

I am a 7 letter city.

234 is a bird
61 is cool
1274 is a part of the face
4713 is a way of saying good

[rajraj]

Simple one.

I am a 7 letter city.

234 is a bird
61 is cool
1274 is a part of the face
4713 is a way of saying good


C H E N N A I

Start with the nearest ! :lol:

[Benny Lava]

It is the only square of a square (4th power) that can be expressed as a sum of squares ! :) I think that was what the book I browsed said! :)

3^4 = 8^2 + 4^2 + 1^2


If I run across the book again I will verify ! :)

May be somebody can write a script (program) to verify the assertion! :)

Benny Lava : Good job! You get an A ! :lol:

Chumma 5 series expand pannadhukellam A kodutha epdi! :lol:

Anyway, I found that 4^4 can be expressed as 1 + 9 + 25 + 100 + 121... so there seems to be at least one another number having the same property? :confused2:

[Benny Lava]

A Clock repairman got an urgent request to fix the broken hands of a clock. When he arrived, it was already getting dark however he fixed the clock hastily. While setting the hands, he referred to his pocket watch. It was six o'clock, so he set the big hand at 12 and the little hand at 6. When he returned soon he got a call saying that the clock is showing wrong time. Surprised, he hurried back to the client's house. He found the clock showing not much past eight. He handed his watch to the client, saying: "Check the time, please. Your clock is not off even by 1 second."

The client had to agree.

Early the next morning the client telephoned to say that the clock hands, apparently gone berserk, were moving around the clock at will. When he rushed over, the clock showed a little past seven. After checking with his watch, the apprentice got angry:
"You are making fun of me! Your clock shows the right time!"

But the mistake was actually on the part of the repairman. What could have gone wrong?

[rajraj]

Anyway, I found that 4^4 can be expressed as 1 + 9 + 25 + 100 + 121... so there seems to be at least one another number having the same property? :confused2:

I will change that to A+ ! :lol: If I find the book I will let you know the name of the author and the publisher. You can write to them! :)

[GP]

Could not find that Lava, leaving it to Rajraj master.

One from me.

All the nine digits are arranged here so as to form four square numbers.
9, 81, 324, 576
How would you put them together so as to form a single smallest possible square number and a single largest possible square number?

[rajraj]

Could not find that Lava, leaving it to Rajraj master.


GP: Solving that requires some knowledge of mechanical clocks, the gear and hands. He has given some clues - 6 PM, pocket watch, erratic etc! Try again or let him give the solution.

I don't think many in your generation ever took a mechanical clock apart! It is all digital these days ! :lol:

[Benny Lava]

Could not find that Lava, leaving it to Rajraj master.


GP: Solving that requires some knowledge of mechanical clocks, the gear and hands. He has given some clues - 6 PM, pocket watch, erratic etc! Try again or let him give the solution.

I don't think many in your generation ever took a mechanical clock apart! It is all digital these days ! :lol:

:lol: No! this has got nothing to do with the knowledge of mechanical clocks, just a simple guess on what could have gone wrong when re-assembling it.

It is not a great deal mathematical either, so everyone can try.

Both the client and repairman are telling the truth, just that on the two occasions when he came to check, the time shown on the clock happens to be the correct time!

[Benny Lava]

Could not find that Lava, leaving it to Rajraj master.

One from me.

All the nine digits are arranged here so as to form four square numbers.
9, 81, 324, 576
How would you put them together so as to form a single smallest possible square number and a single largest possible square number?

GP,

There is no way I can arrive at the answer using pen and paper! However I thought this would be a good opportunity to get into Modeling :lol2:

So I tried model the solution using Microsoft Solver Foundation and here is what I got:

11826^2 = 139854276
30384^2 = 923187456

Below is the OML(optimization modeling language) model I wrote to arrive at this:


Model
[
Decisions[
Integers[1,9],a,b,c,d,e,f,g,h,i
],
Decisions[
Integer[10000,31622],y
],

Constraints[
a + b*10 + c*100 + d*1000 + e*10000 + f*100000 + g*1000000 + h*10000000 + i*10000000 == y*y
Unequal[a,b,c,d,e,f,g,h,i],
],

Goals[
Minimize[a + b*10 + c*100 + d*1000 + e*10000 + f*100000 + g*1000000 + h*10000000 + i*100000000]
]

]

I think this is one hellova tool from Microsoft! :2thumbsup: :2thumbsup: Very simple and easy to solve optimization problems using this. Others interested may try this from here:

http://code.msdn.microsoft.com/solverfoundation

[NOV]

what! still not answered? :roll:
the key is that both times he was called, it was a little after 8 and a little after 7.
if you had said that the time was 8.10 and 7.05, maybe you would have got the answer by now. :D

after all a completely spoilt clock also shows correct time twice a day. :lol2:

[Benny Lava]

what! still not answered? :roll:
the key is that both times he was called, it was a little after 8 and a little after 7.
if you had said that the time was 8.10 and 7.05, maybe you would have got the answer by now. :D

after all a completely spoilt clock also shows correct time twice a day. :lol2:

But this partially clock would show the correct time 24 times a day :wink:

[kirukan]

11826^2 = 139854276
30384^2 = 923187456

Below is the OML(optimization modeling language) model I wrote to arrive at this:


Model
[
Decisions[
Integers[1,9],a,b,c,d,e,f,g,h,i
],
Decisions[
Integer[10000,31622],y
],

Constraints[
a + b*10 + c*100 + d*1000 + e*10000 + f*100000 + g*1000000 + h*10000000 + i*10000000 == y*y
Unequal[a,b,c,d,e,f,g,h,i],
],

Goals[
Minimize[a + b*10 + c*100 + d*1000 + e*10000 + f*100000 + g*1000000 + h*10000000 + i*100000000]
]

]

I think this is one hellova tool from Microsoft! :2thumbsup: :2thumbsup: Very simple and easy to solve optimization problems using this. Others interested may try this from here:

http://code.msdn.microsoft.com/solverfoundation

Kirukans kirukkal code :wink: for getting the same result
--TSQL MSSQL2005


declare @num bigint ,@min bigint ,@max bigint , @sqr bigint,@total int,@rnd varchar(9),@con varchar(9),@cnt int
set @sqr=0

set @num=10000
set @min=0
set @max=0
while(@num<100000)
begin
set @num=@num+1
set @sqr=power(@num,2)
set @rnd= ltrim(str(@sqr))
if(len(@rnd)>9) break
set @total=1
if (charindex('0',@rnd)>0) continue
set @con=''
set @cnt=1
while (@cnt<10)
begin
if(charindex(substring(@rnd,@cnt,1),@con)>0)
begin
set @total=0
break
end
else
set @con=@con+substring(@rnd,@cnt,1)
set @cnt=@cnt+1
end
if(@total=1)
begin
if(@min=0) set @min=@num
set @max=@num
end
end
select @min as Minimum,convert(bigint,power(@min,2)) Min ,@max as Maximum,convert(bigint,power(@max,2)) Max

-

Kirukkan

[rajraj]

I don't know any of these languages/codes! :lol: Please give me a program in FORTRAN,APL or assembly language in 1960s/70s mainframes ! :lol: :lol: :lol:

[kirukan]

I don't know any of these languages/codes! :lol: Please give me a program in FORTRAN,APL or assembly language in 1960s/70s mainframes ! :lol: :lol: :lol:

For you fortran uncompiled



Program PUZZLE
INTEGER Num, min, max,total,cnt,sqr
CHARACTER(LEN=9) rnd, con
sqr=0
num=10000
min=0
max=0
do while(num<100000)
num=num+1
sqr=num*num
Write( rnd, '(i10)' ) sqr
if(len(rnd)>9) break
total=1
if (index('0',rnd)>0) continue
con=''
cnt=1
do while (cnt<10)
if(index(rnd(cnt,1),con)>0) then
total=0
break
else
con=con//rnd(cnt,1)
end if
cnt=cnt+1
end do
if(total=1) then
if(min=0)then
min=num
end if
max=num
end if
end do
print *, min
print *, max
End program

[rajraj]

kirukan: You get an A+ ! :) I don't have access to a mainframe to verify your program. I don't have a teaching assistant to do it! Still, you get an A+ ! :lol:

[rajraj]

GP has enough clues to fix the clock! :)

Here is a simple one:

Which is the odd number here?

25 35 48 63 80

[NM]

is it 25?
everythign else goes by the rule of (n+1) x (n+3) :oops: vendikkai innum saapadanum-pOla

[rajraj]

vendikkai innum saapadanum-pOla

Yes! Lots of it! Try different reasoning ! :lol:

[rajraj]

NM: (n+1)x(n+3) = [(n+2) -1] x[(n+2) +1 ]
= [ (n+2)^2 - 1 ]

Does this give you a clue? :)

[NM]

NM: (n+1)x(n+3) = [(n+2) -1] x[(n+2) +1 ]
= [ (n+2)^2 - 1 ]

Does this give you a clue? :)
hi uncle -
are you saying its wrong?
I thought the sequence should be
3, 8, 15, 24, 35, 48, 63,80, 99, 120 etc......
so the number that's odd is 25 :(

[rajraj]

NM: (n+1)x(n+3) = [(n+2) -1] x[(n+2) +1 ]
= [ (n+2)^2 - 1 ]

Does this give you a clue? :)
hi uncle -
are you saying its wrong?
I thought the sequence should be
3, 8, 15, 24, 35, 48, 63,80, 99, 120 etc......
so the number that's odd is 25 :(


I did not say it was wrong! I asked for different reasoning! You have it in the series you gave! What is common to those numbers? :) To make it plain what is the relationship between 25 and 24 ? :)

[NM]

oooooh! appadiya - its always (the perfect square of a number - 1) .... sorry - didn't quite understand was was required.

(0+2)^2 - 1 = 3
(1+2)^2 - 1 = 8
(2+2)^2 - 1 = 15 etc

ippO naan elementary-thaana illa foundtaion-a? :lol: :lol:

[rajraj]

ippO naan elementary-thaana illa foundtaion-a? :lol: :lol:

I will tell you after ten more puzzles! :lol: A professor has to give enough tests before awarding a grade ! :lol:

You always try for a simple solution ! :)

[rajraj]

Another simple one with numbers ( 4x4 matrix of 16 numbers):


3 7 2 6

4 1 5 8

6 4 5 3

5 6 x 1


What is x ?

[GP]

6.
Both row-wise and column-wise the sum is 18.
ivlO semipleA kuduthirukka vEndaam.
naan play school ku eligibleA?

[rajraj]

ivlO semipleA kuduthirukka vEndaam.


Stress relief ! :lol:

[GP]

For the clock puzzle,

The repairman has put minute hand on the place of seconds hand.

[GP]

Another simple one for stress relief. :wink:

How can you split Rs. 1.15 in six denominations so that you can not have change for one rupee, 50 paise, 25 paise, 10 paise and 5 paise?

[rajraj]

I leave this to those in India to solve! :) I have not handled Indian coins in a long time! I don't even remember what denominations are available now ! :)

[GP]

5 Paise, 10 paise, 20 paise, 25 paise, 50 paise and one rupee denominations can be used. (1 Rupee = 100 paise)

[kirukan]

50+25+10+10+10+10?

[GP]

:clap: Kirukkan :thumbsup:

[Benny Lava]

11826^2 = 139854276
30384^2 = 923187456

Below is the OML(optimization modeling language) model I wrote to arrive at this:


Model
[
Decisions[
Integers[1,9],a,b,c,d,e,f,g,h,i
],
Decisions[
Integer[10000,31622],y
],

Constraints[
a + b*10 + c*100 + d*1000 + e*10000 + f*100000 + g*1000000 + h*10000000 + i*10000000 == y*y
Unequal[a,b,c,d,e,f,g,h,i],
],

Goals[
Minimize[a + b*10 + c*100 + d*1000 + e*10000 + f*100000 + g*1000000 + h*10000000 + i*100000000]
]

]

I think this is one hellova tool from Microsoft! :2thumbsup: :2thumbsup: Very simple and easy to solve optimization problems using this. Others interested may try this from here:

http://code.msdn.microsoft.com/solverfoundation

Kirukans kirukkal code :wink: for getting the same result
--TSQL MSSQL2005


declare @num bigint ,@min bigint ,@max bigint , @sqr bigint,@total int,@rnd varchar(9),@con varchar(9),@cnt int
set @sqr=0

set @num=10000
set @min=0
set @max=0
while(@num<100000)
begin
set @num=@num+1
set @sqr=power(@num,2)
set @rnd= ltrim(str(@sqr))
if(len(@rnd)>9) break
set @total=1
if (charindex('0',@rnd)>0) continue
set @con=''
set @cnt=1
while (@cnt<10)
begin
if(charindex(substring(@rnd,@cnt,1),@con)>0)
begin
set @total=0
break
end
else
set @con=@con+substring(@rnd,@cnt,1)
set @cnt=@cnt+1
end
if(@total=1)
begin
if(@min=0) set @min=@num
set @max=@num
end
end
select @min as Minimum,convert(bigint,power(@min,2)) Min ,@max as Maximum,convert(bigint,power(@max,2)) Max

-

Kirukkan

Splendid! :clap:

I am no good at programming, just script some timepass kirukkals but you call yours a kirukkal!! :shock: You have coded everything from scratch!

Anyway this must be much quicker since we just have to square 9000 numbers and check if they have unique digit, whereas my method would have to undergo at least 9! iterations to arrive at solution! :thumbsup:

[GP]

For the clock puzzle,

The repairman has put minute hand on the place of seconds hand. idhu correct thaana?

[Benny Lava]

For the clock puzzle,

The repairman has put minute hand on the place of seconds hand. idhu correct thaana?

Yes.. yes! Forgot to reply.. Good work! :2thumbsup:

[Benny Lava]

This one is pretty simple!

"A half is a third of it. What is it?" :P

[GP]

1.5.
half is one third of 1.5.
:huh:

[NOV]

idhu correct thaana?Karnan film dialogue, Krishna to Arjuna: "naan thaan konnEn, naandhaan konnEn!" :lol2:

[GP]

idhu correct thaana?Karnan film dialogue, Krishna to Arjuna: "naan thaan konnEn, naandhaan konnEn!" :lol2: :shock: comparing me to Lord Krishna.

[Benny Lava]

1.5.
half is one third of 1.5.
:huh:

Yes! Avlo thaan.. :)

Try this one.

You have 180 pounds of gravel and a balance. You need to weigh out 150 pounds and 30 pounds of gravel. You only have 6 pound and 9 pound standard weights. How would you do that?

[NOV]

idhu correct thaana?Karnan film dialogue, Krishna to Arjuna: "naan thaan konnEn, naandhaan konnEn!" :lol2: :shock: comparing me to Lord Krishna.:P
It was Arjuna who said that dialogue with Krishna mimicking him. :P
Karnan died earlier; arjuna's last arrow just confirmed his death. :P

[GP]

1.5.
half is one third of 1.5.
:huh:

Yes! Avlo thaan.. :)

Try this one.

You have 180 pounds of gravel and a balance. You need to weigh out 150 pounds and 30 pounds of gravel. You only have 6 pound and 9 pound standard weights. How would you do that? any restriction on number of times the balance can be used or maximum weight the balance can hold?

[GP]

Weigh 30 pounds of Gravel in 2 attempts using 15 pound (6+9) standard weight. Remaining should be 150 pounds. :|

[GP]

Should we also check whether the remaining quantity has 150 pounds of Gravel?
if yes,
1) weigh 45 pounds gravel using 15 pound standard weight and 30 pound gravel (that was already weighed). We now have 15 pound standard weight, 30 pound gravel, 45 pound gravel.
2) Weigh 90 pound gravel using 15 pound standard weight, 30 pound gravel, 45 pound gravel. We now have 15 pound standard weight, 30 pound gravel, 45 pound gravel, 90 pound gravel.
3) Check whether the remaining gravel weighs 15 pound using 15 pound standard weight.

2 steps to get 30 pound gravel and 5 steps to confirm whether the remaining gravel weighs 150 pound.

[rajraj]

Too much of stress relief ! :)

Try this one for a change:
You are given a 4x4 matrix of 16 dots. How can you connect them using maximum number of lines without crossing or retracing a line already drawn without lifting your pen? A line is a vertical or horizontal connection between two adjacent dots. No diagonal lines!

[Benny Lava]

Weigh 30 pounds of Gravel in 2 attempts using 15 pound (6+9) standard weight. Remaining should be 150 pounds. :|

Ohh.. sorry.. I forgot to mention that limitation is 3 attempts.

Anyway, your answer is right. Now how can we separate 180 pounds of gravel into 40 pounds and 140 pounds using just 1 pound and 4 pounds standard measure? :D

[Benny Lava]

Too much of stress relief ! :)

Try this one for a change:
You are given a 4x4 matrix of 16 dots. How can you connect them using maximum number of lines without crossing or retracing a line already drawn without lifting your pen? A line is a vertical or horizontal connection between two adjacent dots. No diagonal lines!

The answer must be 15 and the solution seems to be easy, unless I am missing something here :confused2:

[rajraj]

Too much of stress relief ! :)

Try this one for a change:
You are given a 4x4 matrix of 16 dots. How can you connect them using maximum number of lines without crossing or retracing a line already drawn without lifting your pen? A line is a vertical or horizontal connection between two adjacent dots. No diagonal lines!

The answer must be 15 and the solution seems to be easy, unless I am missing something here :confused2:

Maximum number of lines? :) Your solution is the number of lines in a spiral connection! :)

[rajraj]

Ohh.. sorry.. I forgot to mention that limitation is 3 attempts.

Anyway, your answer is right. Now how can we separate 180 pounds of gravel into 40 pounds and 140 pounds using just 1 pound and 4 pounds standard measure? :D

Assuming you have an old style scale with two pans:

180

90 90 (divide equally)

90 45 45 (divide 90 lb equally)

90 45 5 40 (weigh 5 lb from one 45 lb heap)

90+45+5 40

[rajraj]

Be a little more clever! :)







90 45 45 (divide 90 lb equally)

90 45 45+5 (add the 5 lb weight to one pan)
90 45 5 40+5 (remove 5lb material leaving the weights in the pan)
90 45 5 40 (remove the 5 lb weight from the pan)

90+45+5 40

[GP]

Too much of stress relief ! :)

Try this one for a change:
You are given a 4x4 matrix of 16 dots. How can you connect them using maximum number of lines without crossing or retracing a line already drawn without lifting your pen? A line is a vertical or horizontal connection between two adjacent dots. No diagonal lines!

The answer must be 15 and the solution seems to be easy, unless I am missing something here :confused2:

Maximum number of lines? :) Your solution is the number of lines in a spiral connection! :)
17 :huh:
lets number the 16 dots as
01-02-03-04
05-06-07-08
09-10-11-12
13-14-15-16

Connection like this 02-06-05-01-02-03-04-08-12-16-15-14-13-09-10-11-07-03

I hope better solution exist.

[rajraj]

17 :huh:


I hope better solution exist.

Try again ! I will give one for stress relief later ! :lol:

[Benny Lava]

Be a little more clever! :)







90 45 45 (divide 90 lb equally)

90 45 45+5 (add the 5 lb weight to one pan)
90 45 5 40+5 (remove 5lb material leaving the weights in the pan)
90 45 5 40 (remove the 5 lb weight from the pan)

90+45+5 40

Excellent once again! :clap:

[Benny Lava]

Too much of stress relief ! :)

Try this one for a change:
You are given a 4x4 matrix of 16 dots. How can you connect them using maximum number of lines without crossing or retracing a line already drawn without lifting your pen? A line is a vertical or horizontal connection between two adjacent dots. No diagonal lines!

The answer must be 15 and the solution seems to be easy, unless I am missing something here :confused2:

Maximum number of lines? :) Your solution is the number of lines in a spiral connection! :)

Oh, sorry! When you said "without crossing" I thought it meant that only two edges can be drawn from a node.

And for your question, it can be solved easily if we apply Graph theory. The requirement is basically to construct an Euler's path through 16 nodes (number of edges are indeterminate at the moment).

For an Euler's path to exist, the graph must have zero or two nodes with odd number of edges. The maximum number of edges for a nxn matrix without considering Euler's path would be 2*n*(n-1). In this case it is 24. In such a setup the node properties is as such:

4 nodes with 2 edges
8 nodes with 3 edges
4 nodes with 4 edges

Clearly, the 8 nodes with 3 edges bothers us since we can not have more than 2 nodes as such. This means 6 nodes have to shed 1 edge each. If we want to maximize the number of edges, this can be accomplished by shedding merely 3 edges (shared by 6 nodes as a pair).

So the final solution is 24 - 3 = 21 edges.

Diagramatically, the 3 edges to remove are the middle edges of any three exterior sides. Like:

(1,2) to (1,3) &
(2,4) to (3,4) &
(4,2) to (4,3)

and other symmetrical variation of the same topology.

[GP]

yellAm ok, yeppadi andha 21 lines connect pannuveengannu sollunga. :?

[kirukan]

2-1-5-6-2-3-7-6-10-9-13-14-10-11-15-16-12-11-7-8-4-3.
Kirukan kirukkiya kodugal 21

is it right?

[GP]

:clap: :clap: kirukan

[Benny Lava]

yellAm ok, yeppadi andha 21 lines connect pannuveengannu sollunga. :?

Sorry, lunch hour la office lerndhu post pannadhaala answer complete panna mudila.. anyway the answer should be easy once you know what shape you are going to get, which I described in my previous post.

There are several routes for this, the one which I jotted down in my paper is:

1-2-6-7-3-4-8-7-11-12-16-15-11-10-14-13-9-10-6-5-1

[rajraj]

21 is correct. There is more than one solution! :) This is an IQ test problem. Those who take it have no clue about graph theory ! :)

Thanks Benny Lava for taking me back to the days when I taught some graph theory and even used it ( isomorphic graphs) in chip design ! :)

[rajraj]

Stress relief (?) !

Try connecting those 16 dots in a 4x4 matrix with minimum number of lines! I am sure you did it with 9 dots in a 3x3 matrix! :)

[kirukan]

15?

1-2-3-4-8-12-16-15-14-13-12-9-5-6-7-11-10

[rajraj]

15?

1-2-3-4-8-12-16-15-14-13-12-9-5-6-7-11-10

No! You have to think outside the box as you did with 9 dot matrix. There is no vertical and horizontal line restriction! :)

[Benny Lava]

15?

1-2-3-4-8-12-16-15-14-13-12-9-5-6-7-11-10

No! You have to think outside the box as you did with 9 dot matrix. There is no vertical and horizontal line restriction! :)

Six lines is the least I could think of. Is that the answer?

[Benny Lava]

21 is correct. There is more than one solution! :) This is an IQ test problem. Those who take it have no clue about graph theory ! :)

:lol: Sorry, Konigsberg bridge problem was the first thing that came to my mind when I understood your question. Naturally I ended up using it's solution as an analogue.

[Benny Lava]

[tscii:2a1b405e12]I hope I haven't posted this one before. Below is a minor modification of the concept of crossword puzzles. Instead of guessing the words, you will have to guess numbers to fill up the blank cells. There exist only one unique solution to this puzzle which can be arrived at purely using common sense. To avoid Googling, the names of the characters in clues have been changed. Please do not Google for answer.

Below are the conversion table for those who are not familiar with the units:

20 shillings = 1 pound sterling
1 acre = 4840 square yards
1 rood = ¼ acre
1 mile = 1760 yards

The puzzle was composed in early 20th century, so keep that in mind when estimating years. I found it pretty enjoyable and challenging too. So have fun :wink:


##################################################
# 1 # # 2 # 3 # ######## 4 #
# # # # # ######## #
# # # # # ######## #
# # # # # ######## #
# # # # # ######## #
# # # # # ######## #
##################################################
# ######## # 5 ######## 6 # #
# ######## # ######## # #
# ######## # ######## # #
# ######## # ######## # #
# ######## # ######## # #
# ######## # ######## # #
##################################################
# ######## ######## 7 # # #
# ######## ######## # # #
# ######## ######## # # #
# ######## ######## # # #
# ######## ######## # # #
# ######## ######## # # #
##################################################
######## 8 # # 9 # ###############
######## # # # ###############
######## # # # ###############
######## # # # ###############
######## # # # ###############
######## # # # ###############
##################################################
# 10 # ######## 11 # # 12 # 13 #
# # ######## # # # #
# # ######## # # # #
# # ######## # # # #
# # ######## # # # #
# # ######## # # # #
##################################################
# ###################### 14 # # #
# ###################### # # #
# ###################### # # #
# ###################### # # #
# ###################### # # #
# ###################### # # #
##################################################
# 15 # ######## 16 # ######## #
# # ######## # ######## #
# # ######## # ######## #
# # ######## # ######## #
# # ######## # ######## #
# # ######## # ######## #
##################################################


ACROSS
1. Area in square yards of Pig Farm.
5. Age of Farmer Ronnie’s daughter,
Celia.
6. Difference in yards between the length
and breath of Pig Farm.
7. Number of roods in Pig Farm
multiplied by 9 Down.
8. Date (A.D.) when the Ronnie family
originally bought Little Pigley.
10. Farmer Ronnie’s age.
11. The year of birth of Melissa, Farmer
Ronnie’s youngest.
14. The perimeter in yards of Dog’s
Mead.
15. The cube of Farmer Ronnie’s walking
speed in miles per hour.
16. 15 Across minus 9 Down.

DOWN
1. The value in shillings per acre of
Pig Farm.
2. The square of Mrs. Howard’s (Farmer
Ronnie’s mother-in-law’s) age.
3. Age of Melissa (see 11 Across).
4. Value of Pig Farm in pounds
sterling.
6. The age of Farmer Ronnie’s first-born,
Ted, who will be twice as old as Melissa
next year.
7. The square of the number of yards in
the breadth of Pig Farm.
8. The number of minutes Farmer Ronnie
takes to walk one and a third times
around Pig Farm.
9. See 10 Down.
10. 10 Across times 9 Down.
12. One more than the sum of the digits
in the row marked by the arrows.
13. The length of tenure in years of Little
Pigley by the Ronnie family.

[/tscii:2a1b405e12]

[rajraj]

Six lines is the least I could think of. Is that the answer?

7 is the answer given by the author of the book from which I took the problem. He asserts that it can not be done with less than 7 lines. I don't know whether there is a later edition of the book with 6 as the answer.

Please provide me a description of the lines with node numbers.

Thanks! :)

[Benny Lava]

Six lines is the least I could think of. Is that the answer?

7 is the answer given by the author of the book from which I took the problem. He asserts that it can not be done with less than 7 lines. I don't know whether there is a later edition of the book with 6 as the answer.

Please provide me a description of the lines with node numbers.

Thanks! :)

Cartesian co-ordinates for the solution would be like:

0,3
4,3
-1,0
4,0
0,4
0,1
2,1

The 16 grid points are between (0,0) to (3,3)

[rajraj]

Cartesian co-ordinates for the solution would be like:

0,3
4,3
-1,0
4,0
0,4
0,1
2,1

The 16 grid points are between (0,0) to (3,3)

Thanks! I have to draw it to check! :)

[rajraj]

Bebnny Lava: Please check the coordinates! :)

[GP]

Benny Lava, you have altered the Dogsmead puzzle right?
It was posted long time back in this thread. After almost a month of try I had put the answer.
Not in the mood to try it again.

[rajraj]

GP: Please verify whether this one is repeated. In that case I won't try. Even otherwise this puzzle gives a lot of clues ! :lol:


Dogsmead? (http://forumhub.mayyam.com/hub/viewtopic.php?p=1647039#1647039)

[GP]

yes master. it is the same.
BTW, is Benny Lava = thamizhvaanan?
Both have given the same puzzle and have same avatar? :roll:

[Benny Lava]

Benny Lava, you have altered the Dogsmead puzzle right?
It was posted long time back in this thread. After almost a month of try I had put the answer.
Not in the mood to try it again.

Oh yeah! So my memory was right! :P

I was too lazy to check whether I have already posted it :lol:

Anyway, people who haven't tried it can give a fresh try.

But I see that in your solution farmer Dunk's daughter is older than him.. when are you going to fix that? :twisted: It is impossible unless of course if she had a Time travel machine :lol2:

[Benny Lava]

Bebnny Lava: Please check the coordinates! :)

I checked again :roll:

The 4 x 4 matrix is formed by the co-ordinates 0,0 to 3,3. How can I draw it here? :(

[GP]

BTW previous puzzle thread 100 pages mudinju indha thread aarambichirukeenga nu nenachEn.
but that thread has just 6 pages. adhukkulla avasarapattu pudhu thread aarambichadhu yaaru?

[GP]

Benny Lava, you have altered the Dogsmead puzzle right?
It was posted long time back in this thread. After almost a month of try I had put the answer.
Not in the mood to try it again.
But I see that in your solution farmer Dunk's daughter is older than him.. when are you going to fix that? :twisted: It is impossible unless of course if she had a Time travel machine :lol2:
Diwali kulla fix panrEn, adhu varaikkum yaarum answer sollaama irundhaa.

[kirukan]

15?

1-2-3-4-8-12-16-15-14-13-12-9-5-6-7-11-10

No! You have to think outside the box as you did with 9 dot matrix. There is no vertical and horizontal line restriction! :)

Then its 7
1-2-3-4
4-8-12-16
16-15-14-13
13-9-5
5-6-7
7-11
11-10

Is it right r still i have to go out of box.

[Benny Lava]

Bebnny Lava: Please check the coordinates! :)

I checked again :roll:

The 4 x 4 matrix is formed by the co-ordinates 0,0 to 3,3. How can I draw it here? :(

I tried to plot it in excel. Please check the screen shot below:

http://i34.tinypic.com/2w21quo.png

Start from lower right corner of the 4x4 grid.

[rajraj]

Is it right r still i have to go out of box.

Correct! If you went out of the box(grid) you will get another solution! :)

[rajraj]

I tried to plot it in excel. Please check the screen shot below:

http://i34.tinypic.com/2w21quo.png

Start from lower right corner of the 4x4 grid.

Your solution crosses a line already drawn which is not permitted as stated in the very first connection puzzle. Try again ! :)

[rajraj]

Let us move from squares to circles and balls! :)

A 40 cm diameter ball is on the floor touching a wall leaving a gap where the perpendicular wall joins the floor. Can a ball with 7 cm diameter pass through the gap (between the ball, the floor and the wall) ?

[GP]

Radius of ball 20cm.

According to Pythagoras,
Distance between centre of ball and the edge of wall = 28.28 cm (sqrt(20^2+20^2)).
Of this 28.28cm, 20cm is already covered by the ball.

So a ball with diameter less than 8.28cm can pass through the gap.

Was a nice stress relief.

http://i38.tinypic.com/jazw2v.jpg

[Benny Lava]

I tried to plot it in excel. Please check the screen shot below:

http://i34.tinypic.com/2w21quo.png

Start from lower right corner of the 4x4 grid.

Your solution crosses a line already drawn which is not permitted as stated in the very first connection puzzle. Try again ! :)

Oh! I didn't know that we shouldn't cross any line.. but then, with that constraint even the 9 dots puzzle has no solution.. right? :)

[rajraj]

Oh! I didn't know that we shouldn't cross any line.. but then, with that constraint even the 9 dots puzzle has no solution.. right? :)

If a line goes through a dot it is not considered crossing a line. If it crosses between dots it is considered crossing a line! :) You could try something similar to the solution for 9 dot puzzle ! :)

[GP]

Raj master, ball puzzle answer panni rukkEn.
paarthuttu correcttA nu sollunga.

[rajraj]

Radius of ball 20cm.

According to Pythagoras,
Distance between centre of ball and the edge of wall = 28.28 cm (sqrt(20^2+20^2)).
Of this 28.28cm, 20cm is already covered by the ball.

So a ball with diameter less than 8.28cm can pass through the gap.

Was a nice stress relief.

http://i38.tinypic.com/jazw2v.jpg

8.28 cm is the distance between the large ball and the corner where the wall meets the floor. What is the distance between that corner and the largest smaller ball that can go into that gap? :)

I am sorry I dd not scroll back to see any solutions posted ! :)

[Benny Lava]

Oh! I didn't know that we shouldn't cross any line.. but then, with that constraint even the 9 dots puzzle has no solution.. right? :)

If a line goes through a dot it is not considered crossing a line. If it crosses between dots it is considered crossing a line! :) You could try something similar to the solution for 9 dot puzzle ! :)

I can't figure it out, unless I am allowed to retrace my path along one of the lines. Could you post the solution please?

[rajraj]

I can't figure it out, unless I am allowed to retrace my path along one of the lines. Could you post the solution please?

Expand the grid to (5x5) still retaining only 4x4 dots and shifting them by one unit in both coordinates. Now, you have two points (0,4) and (4,0) which are not dots. The solution contains the line connecting these two points. This line connects three of the off main diagonal dots in the 4x4 dot matrix. It should be easy now! :) The solution does not retrace any line!

[rajraj]

Benny Lava: When you gave your first solution you had (4,0) and (0,4) as two of the points. I thought you had the solution and asked you to check ! :)

[Benny Lava]

I can't figure it out, unless I am allowed to retrace my path along one of the lines. Could you post the solution please?

Expand the grid to (5x5) still retaining only 4x4 dots and shifting them by one unit in both coordinates. Now, you have two points (0,4) and (4,0) which are not dots. The solution contains the line connecting these two points. This line connects three of the off main diagonal dots in the 4x4 dot matrix. It should be easy now! :) The solution does not retrace any line!

Rajraj,

To me it looks impossible. Please post your complete solution.

[rajraj]

Benny Lava,
I think you are trying to modify the solution you gave in the picture. Start all over again with the clue I gave! I am sure you will find it! If not I will post the solution! :)

[Benny Lava]

Benny Lava,
I think you are trying to modify the solution you gave in the picture. Start all over again with the clue I gave! I am sure you will find it! If not I will post the solution! :)

Let me clarify whether my assumptions are right.

1. I am not allowed to retrace my path
2. Paths cannot overlap
3. It is a continuous path, i.e, I have to draw the path without lifting my pen.

Is that so?

[rajraj]

Benny Lava,
I think you are trying to modify the solution you gave in the picture. Start all over again with the clue I gave! I am sure you will find it! If not I will post the solution! :)

Let me clarify whether my assumptions are right.

1. I am not allowed to retrace my path
2. Paths cannot overlap
3. It is a continuous path, i.e, I have to draw the path without lifting my pen.

Is that so?

Yes! Also you can not cross a line except at a dot ! :) When I said 'outside the box' to kirukan I meant going outside the 4x4 dot matrix ! :)

[Benny Lava]

Benny Lava,
I think you are trying to modify the solution you gave in the picture. Start all over again with the clue I gave! I am sure you will find it! If not I will post the solution! :)

Let me clarify whether my assumptions are right.

1. I am not allowed to retrace my path
2. Paths cannot overlap
3. It is a continuous path, i.e, I have to draw the path without lifting my pen.

Is that so?

Yes! Also you can not cross a line except at a dot ! :) When I said 'outside the box' to kirukan I meant going outside the 4x4 dot matrix ! :)

Andha naalavadhu point thaaney idikudhu :lol:

Anyway, my reasoning goes like this:
1. The first stroke you described (4,0) to (0,4) covers the 3 off-diagonal dots.
2. Since there are no intersections, drawing lines along both off-diagonals parallel to the anti-diagonal is ruled out but drawing the anti-diagonal is still a possibility. From here let us branch into two possibilities:
a) We draw an anti-diagonal: Now we cannot draw a main diagonal since it would intersect. Now consider the two middle points in main diagonal (1,2) & (2,1). To cover these points we require no less than 4 lines which leaves us only one line to cover other dots, definitely some would be left off.
b) We do not draw an anti-diagonal: To cover the same two middle points, we can only use vertical lines or connect by drawing main diagonal (lines perpendicular to it would intersect) Both the cases would have already consumed 4 strokes. It is impossible for the last stroke to cover rest of the dots.

So it seems like it is possible to connect the dots by just using 6 lines with above constraints. Am I missing something here? :oops:

[rajraj]

. --o--- o----o---o <---- Start here at (4,4)

.....o.....o.....o.....o

.....o.....o.....o.....o

.....o.....o.....o.....o

..........................<--- from (0,4) draw a diagonal line to (4,0)


You have drawn two lines so far connecting 7 dots!
Now use only vertical and horizontal lines to complete the solution! :) ( Remember crossing a line at a dot is permissible! )

This vaathiyaar is very patient but a little bit demanding ! :lol: Not a great artist ! :)

[GP]

Radius of ball 20cm.

According to Pythagoras,
Distance between centre of ball and the edge of wall = 28.28 cm (sqrt(20^2+20^2)).
Of this 28.28cm, 20cm is already covered by the ball.

So a ball with diameter less than 8.28cm can pass through the gap.

Was a nice stress relief.

http://i38.tinypic.com/jazw2v.jpg

8.28 cm is the distance between the large ball and the corner where the wall meets the floor. What is the distance between that corner and the largest smaller ball that can go into that gap? :)

I am sorry I dd not scroll back to see any solutions posted ! :) Should be the same? A ball with diameter less than 8.28cm can pass through the gap. :huh:
Or should I think out of the box? :think:

[rajraj]

Should be the same? A ball with diameter less than 8.28cm can pass through the gap. :huh:
Or should I think out of the box? :think:

If a large ball left a gap near the corner for a smaller ball, the smaller ball will also leave a gap near the corner for a still smaller ball ! :)

Now you don't have to think outside the box! The equations get a little more complex ! A little more stress ! :)

[GP]

Should be the same? A ball with diameter less than 8.28cm can pass through the gap. :huh:
Or should I think out of the box? :think:

If a large ball left a gap near the corner for a smaller ball, the smaller ball will also leave a gap near the corner for a still smaller ball ! :)

Now you don't have to think outside the box! The equations get a little more complex ! A little more stress ! :) but ur question was whether a 7cm dia ball can pass through the gap. the answer is yes. What is the problem if there's still a gap. :huh:

[rajraj]

but ur question was whether a 7cm dia ball can pass through the gap. the answer is yes. What is the problem if there's still a gap. :huh:

Now, you have to think outside the box ! :) Extend the line connecting the corner to the center of the large 40 cm diameter ball to the surface of the ball. The length of that line is 48.28 cm. But the diameter of the ball is 40 cm. If similar length is 8.28 cm what will be diameter of the largest ball? In other words you have to find out the diameter of the largest ball that can pass through that gap. Only then you can make a conclusion. Try 8.25 cm diameter ball. You will have to make a dent in the wall and the floor! :) 8.25 < 8.28 ! :)

[GP]

oops, technical fault.
7cm dia ball cant pass through the gap.
A Ball with dia less than 6.86cm can pass through the gap.

[rajraj]

GP: :)

Let us move to triangles:

Two right angled triangles have a common hypotenuse (joined at the hypotenuse of same length). The other two sides of one triangle are 18 and 13 units. The other two sides of the second triangle are 20 and 7 units.
Obviously the units are not decimal (base ten numbers). What is the length of the hypotenuse and what is the base used?

All of you seem to be in IT sector. This should be stress relief for you! :)

[GP]

i yaam in construction sector, but work yellAm computer la thaan. :)

[disk.box]

தசம எண்கள் கிடையாதா? :confused2:
பித்தாகரஸ் தேற்றம் கைகொடுக்கவில்லையே :huh:

[rajraj]

Pythogorean theorem will help to find the base(not decimal)! :) First you have to solve for the unknown base ! :)




தசம எண்கள் கிடையாதா? :confused2:
பித்தாகரஸ் தேற்றம் கைகொடுக்கவில்லையே :huh:

[rajraj]

i yaam in construction sector, but work yellAm computer la thaan. :)

You must be the one who populated Avinasi Road,CBE with all those buildings! :lol:

[GP]

GP: :)

Let us move to triangles:

Two right angled triangles have a common hypotenuse (joined at the hypotenuse of same length). The other two sides of one triangle are 18 and 13 units. The other two sides of the second triangle are 20 and 7 units.
Obviously the units are not decimal (base ten numbers). What is the length of the hypotenuse and what is the base used?

All of you seem to be in IT sector. This should be stress relief for you! :)
If two triangles have common hypotenuse, it should be a rectangle, right (for any base)? :?

[rajraj]

If two triangles have common hypotenuse, it should be a rectangle, right (for any base)? :?

No! Draw a circle with the hypotenuse as the diameter. Pick any point on the circle. The diameter will subtend a right angle ! :) It is known as Thales' theorem!

[Benny Lava]

GP: :)

Let us move to triangles:

Two right angled triangles have a common hypotenuse (joined at the hypotenuse of same length). The other two sides of one triangle are 18 and 13 units. The other two sides of the second triangle are 20 and 7 units.
Obviously the units are not decimal (base ten numbers). What is the length of the hypotenuse and what is the base used?

All of you seem to be in IT sector. This should be stress relief for you! :)

The base of the number system cannot be lesser than 9 (since there is a number with digit 8) at the same time it cannot be greater than 20, because if so all the numbers decimal equivalent would be the same (the biggest number is 20) and we already observe that the numbers don't meet the requirement in decimal base.

A tiny piece of Python code to iterate yields the answer to be base 12:


def toDecimal( something,base ):
index = 0
number = 0
for el in something[ : : -1 ]:
number += int( el ) * ( base**index )
index += 1
return number

a1 = "18"
b1 = "13"

a2 = "20"
b2 = "7"

for i in range(9,20):
print "Trying base %i" %i
if toDecimal(a1,i)**2 + toDecimal(b1,i)**2 == toDecimal(a2,i)**2 + toDecimal(b2,i)**2:
print "Success!!! The base is %i" %i
break


Trying base 9
Trying base 10
Trying base 11
Trying base 12
Success!!! The base is 12

[Benny Lava]

. --o--- o----o---o <---- Start here at (4,4)

.....o.....o.....o.....o

.....o.....o.....o.....o

.....o.....o.....o.....o

..........................<--- from (0,4) draw a diagonal line to (4,0)


You have drawn two lines so far connecting 7 dots!
Now use only vertical and horizontal lines to complete the solution! :) ( Remember crossing a line at a dot is permissible! )

This vaathiyaar is very patient but a little bit demanding ! :lol: Not a great artist ! :)

Rajraj Sir,

Still I am unable to arrive at the solution :oops: Please post the complete co-ordinates.

[rajraj]

The base of the number system cannot be lesser than 9 (since there is a number with digit 8) at the same time it cannot be greater than 20, because if so all the numbers decimal equivalent would be the same (the biggest number is 20) and we already observe that the numbers don't meet the requirement in decimal base.

A tiny piece of Python code to iterate yields the answer to be base 12:


def toDecimal( something,base ):
index = 0
number = 0
for el in something[ : : -1 ]:
number += int( el ) * ( base**index )
index += 1
return number

a1 = "18"
b1 = "13"

a2 = "20"
b2 = "7"

for i in range(9,20):
print "Trying base %i" %i
if toDecimal(a1,i)**2 + toDecimal(b1,i)**2 == toDecimal(a2,i)**2 + toDecimal(b2,i)**2:
print "Success!!! The base is %i" %i
break


Trying base 9
Trying base 10
Trying base 11
Trying base 12
Success!!! The base is 12

Correct! :) You can solve it using paper and pencil. Assume n to be the base ,express the numbers in decimal base and use Pythogorean theorem. You get the equation:

(n+8)^2 + (n+3)^2 = (2n)^2 + 7^2

or

n^2 -11n -12 =0

or

n=12

:)

You get an A- because you used a script ! :lol:

[rajraj]

. --o--- o----o---o <---- Start here at (4,4)

.....o.....o.....o.....o

.....o.....o.....o.....o

.....o.....o.....o.....o

..........................<--- from (0,4) draw a diagonal line to (4,0)


You have drawn two lines so far connecting 7 dots!
Now use only vertical and horizontal lines to complete the solution! :) ( Remember crossing a line at a dot is permissible! )


line 1...................... (4,4) to (0,4) Connects the dots in top row

line 2...................... (0,4) to (4,0) Connects the dots below the main diagonal in the matrix of dots

line 3....................... (4,0) to (4,3) Connects the bottom three dots in the last column

line 4........................ (4,3) to (1,3) Connects the dots in the row below the top row

line 5......................... (1,3) to (1,1) Connects the bottom three dots in the first column

line 6......................... (1,1) to (3,1) Connects the first three dots in the bottom row

line 7..........................(3,1) to (3,2) Connects the bottom two dots in the third column

Benny Lava: I think you missed something in what I said or misunderstood something! :)

May be, I was not very clear ! :(

[Benny Lava]

Correct! :) You can solve it using paper and pencil. Assume n to be the base ,express the numbers in decimal base and use Pythogorean theorem. You get the equation:

(n+8)^2 + (n+3)^2 = (2n)^2 + 7^2

or

n^2 -11n -12 =0

or

n=12

:)

You get an A- because you used a script ! :lol:

Oh! Why didn't I think of that :oops:

[Benny Lava]

line 1...................... (4,4) to (0,4) Connects the dots in top row

line 2...................... (0,4) to (4,0) Connects the dots below the main diagonal in the matrix of dots

line 3....................... (4,0) to (4,3) Connects the bottom three dots in the last column

line 4........................ (4,3) to (1,3) Connects the dots in the row below the top row

line 5......................... (1,3) to (1,1) Connects the bottom three dots in the first column

line 6......................... (1,1) to (3,1) Connects the first three dots in the bottom row

line 7..........................(3,1) to (3,2) Connects the bottom two dots in the third column

Benny Lava: I think you missed something in what I said or misunderstood something! :)

May be, I was not very clear ! :(

Oh.. I was thinking that you were giving a solution with 6 lines. I was pretty convinced that it couldn't be done with any fewer than 7 lines.

No Rajraj sir, you were pretty clear.. it was just an assumption on my part that you were posting a solution for connecting with 6 lines.

OTOH, you said even in the original problem (9 dots version) the lines aren't supposed to cross. The solution that I have crosses each other. Could you please post the solution to that puzzle as well? :)

[rajraj]

Benny Lava : Here is the solution for 3x3 dot matrix ! :)



.....o.....o.....o <- Start here at (3,3)

.....o.....o.....o

.....o.....o.....o

.................................



line 1...................... (3,3) to (0,3) Connects the dots in top row

line 2...................... (0,3) to (3,0) Connects the dots below the main diagonal in the matrix of dots

line 3....................... (3,0) to (3,3) Connects the three dots in the last column making a triangle

line 4........................ (3,3) to (0,0) Connects the dots in the other diagonal

You should have an arrow like figure! :)

[Benny Lava]

Yes... this is what I have seen but in this solution line 2 and line 4 intersects at (1.5,1.5) hence my original confusion. :)

[rajraj]

Yes... this is what I have seen but in this solution line 2 and line 4 intersects at (1.5,1.5) hence my original confusion. :)

You are right! I forgot the restriction and gave you the solution for a simpler puzzle. With the added restriction the solution will resemble the letter Z , the letter S or the letter G requiring a minimum of five lines.

(That tells me that I should not do anything serious beyond midnight and three hours beyond my bed time! :lol: )

[Benny Lava]

Ok :)

[rajraj]

STRESS RELIEF

When I posted the 4x4 dot matrix puzzle I expected soutions with only horizontal and vertical lines! :) Kirukan gave one. There are others. Can you find them? You will be surprised to see how simple they are!

[rajraj]

Benny Lava : Here is the correct solution for 3x3 dot matrix ! :)



.....o.....o.....o <- Start here at (3,3)

.....o.....o.....o

.....o.....o.....o

.................................



line 1...................... (3,3) to (0,3) Connects the dots in top row

line 2...................... (0,3) to (3,0) Connects the dots below the main diagonal in the matrix of dots

line 3....................... (3,0) to (3,2) Connects the bottom two dots in the last column

line 4........................ (3,2) to (1,2) Connects the three dots in the middle row

line 5....................... (1,2) to (1,1) Connects the bottom two dots in the first column

This is similar to the solution for the 4x4 dot matrix.

(There is a pile of small sheets in front of the monitor . I picked the wrong one earlier. I have already given you a few of the other possible solutions with 5 lines! :) )

[rajraj]

I hope all of you visiting this thread had enough stress relief! :)

Here is another problem to use what a Greek mathematician discovered:

A five meters long ladder is leaning against a wall. A cubic box ( 1m x 1m x 1m) is against the wall. One edge of the cube touches the ladder and one side of the cube touches the wall with another side on the ground. That is, the one meter cube box fits snugly in the gap between the ladder and the wall sitting on the ground.

What is the distance of the foot(base) of the ladder from the wall?

[rajraj]

Close to 400 views! But, no solution yet! :) Time to assemble all information to solve the puzzle!

1. Draw a diagram - cross section

2. The ladder leaning against the wall makes a triangle with the wall and the floor as sides. Hypotenuse has a length of 5 meters.

3. Inside the triangel is a square of side 1 meter. The square divides the large triangle into two small triangles.

4. One triangle sits on the square. That is the length of a side is 1 meter

5. The other small triangle is on the side sharing a side with the square. That is one side of the triangle is 1 meter.

6. The bases ( horizontal sides) of the two small triangles are parallel to each other(and parallel to the ground) .

7. The other sides of the small triangles are also parallel( and parallel to the wall)

8. There must be something special about the two small triangles! :)

9. We learnt a few things in geometry and there is Pythogorean theorem!

Now try! :)

Hint: you have to use one unknown - either the horizontal side of the triangle on the side of the square and relate to the vertical side of the other small triangle or vice versa ! :)

[GP]

what, this has not been solved even after 10 days?
this doesn't look that much difficult.

i think 2 answers possible for dis.
4.84m and 1.26m?
just a rough calculation.
if wrong, tell me. will once again try it thinking out of box.

[rajraj]

GP:
Which one makes sense if you are doing something on the wall? It will help others if you explain how you solved it! :)

I hope it is not by trial and error ! :lol:

[GP]

let base of the bigger triangle be 'b' and height be 'h'.
Relating the two small triangles,
(h-1)X(b-1)=1X1 => hb=h+b => h=b/(b-1) --> Eqn I
According to Pythogoras,
h^2+b^2=25 --> Eqn II
Solving Eqn I and II,
(b/(b-1))^2+b^2=25 => b^2(1+1/(b-1)^2)=25
Solving the above eqn we can find the value of 'b' and hence 'h'.
of course i used excel spreadsheet for solving the last eqn.

[rajraj]

Correct! :)

1.260 is the right answer. With the other length you won't need a ladder! If one is the base the other will be the height. You don't need a ladder for a height of 1.26 m ! :)

In case people are wondering equation one is from the fact that the two small triangles are similar! :)

If nobody posts the next puzzle I will post one soon! :)


How do you solve it without writing a script?

[GP]

high school maths la indha diagram padicha nyabagam konjam irundhuchu, especially that relationship between smaller triangles.
idhukku script, screenplay yellAm thEvai padala.
it was a nice stress reliever. :wink:

[rajraj]

of course i used excel spreadsheet for solving the last eqn.

How do you solve it without a spreadsheet/script?

Hint: Let (b-1) = x, the distance between the square and the ladder along the floor. Rewrite the last equation with x. Express the 4th degree equation as a square of a quadratic equation. Now, you won't need a spreadsheet/script! :) A simple $4 calculator will do or an approximation! :)

( These puzzles are from practice IQ tests where a laptop is not allowed! :lol: )

[Benny Lava]

A simple time pass puzzle:

Two taps are filling a bath. The first tap, by itself can fill the bath in one hour if the plug is in. The second tap can fill the bath in 30 minutes if the plug is in. With the plug out and both taps turned off, a full bath will empty in 45 minutes. How long does it take to fill an empty bath with both taps running and the plug out?

[kirukan]

36 mins?
x/30+x/60-x/45=1

[Benny Lava]

Sorry for the late response. Of course the answer is correct! :clap:

Next question:
A race course owner wants to install photoelectric timing device around his race track. The idea is to be able to conduct races & measure time taken for any race course length between 1000 metres to 31000 metres (in multiples of 1000 metres). Assume that he is free to conduct race in either direction of the race course. However the photoelectric timing device turns out to be pretty expensive. What is the minimum number of sensors required for his purpose and how should they be arranged?

[Benny Lava]

Sorry for the late response. Of course the answer is correct! :clap:

Next question:
A race course owner wants to install photoelectric timing device around his race track. The idea is to be able to conduct races & measure time taken for any race course length between 1000 metres to 31000 metres (in multiples of 1000 metres). Assume that he is free to conduct race in either direction of the race course. However the photoelectric timing device turns out to be pretty expensive. What is the minimum number of sensors required for his purpose and how should they be arranged?

No attempts? Guess I should explain the solution a bit more. Say if the length of the circular track is 3000 metres and the owner wants to conduct time tests for distance 1000 m or 2000 m, he can simply install two photoelectric device 1000 m from each other. For 2000 meter measurement he simply has to conduct race in the opposite direction.

[rajraj]

Benny Lava: You have to state the assumptions clearly, unless I misunderstood! I have not been in a race in 60 years! :lol:

Do you need a photo sensor at the starting point?
Is the starting point fixed?

Depending on your answers to these questions the solutions will be different! :)

Hope you had a nice deepaavaLi ! :)



No attempts? Guess I should explain the solution a bit more. Say if the length of the circular track is 3000 metres and the owner wants to conduct time tests for distance 1000 m or 2000 m, he can simply install two photoelectric device 1000 m from each other. For 2000 meter measurement he simply has to conduct race in the opposite direction.

[Benny Lava]

Benny Lava: You have to state the assumptions clearly, unless I misunderstood! I have not been in a race in 60 years! :lol:

Do you need a photo sensor at the starting point?
Is the starting point fixed?

Depending on your answers to these questions the solutions will be different! :)

Hope you had a nice deepaavaLi ! :)



No attempts? Guess I should explain the solution a bit more. Say if the length of the circular track is 3000 metres and the owner wants to conduct time tests for distance 1000 m or 2000 m, he can simply install two photoelectric device 1000 m from each other. For 2000 meter measurement he simply has to conduct race in the opposite direction.

Rajraj sir, Yes you are right. One sensor each at starting and ending point is needed to measure the time taken for each race. The starting point can be changed. The same photo electric sensor can be used as a starting point or an ending point. :)

[rajraj]

More hints:

Take the 3000 meter circular race track, place one sensor at 0 meters( or 3000 meters if you go around) and another sensor at 2000 meters.

For a 1000 meter race start at the first sensor and end at the second sensor.

For a 2000 meter race start at the second sensor(at 1000 meters) and finish at the first sensor(at 3000 meters or 0 meters).

For a 3000 meter race start at the first sensor and finish at the same sensor going around.

You need only two sensors and you do not have to change directions. But you moved the starting point. That is the key - moving the starting point for different lengths!

Hope this helps in the 31000 meter race course problem! :)

[rajraj]

While you are racing to solve the race course puzzle, here is something to relax! :)

On Sunday I watched the movie "A Disappearing Number", video taped version of the drama with the same title produced by the National Theatre(UK). It is about Ramanujan, the mathematician of the millennium. The number 1729 was in the movie. The conversation between Hardy and Ramanujan went somewhat like this when Hardy visited Ramanujan in the hospital.

Hardy: I rode in a taxi with the number 1729. An ordinary number

Ramanujan: No Hardy! It is a special number. It is the smallest number that can be exppressed as.............................

Fill in the blanks!

( Don't search the internet or run to the libray/bookstore for books on Ramanujan! :) )

[rajraj]

Some more words of Ramanujan as hint:

Ramanujan: No Hardy! It is the smallest number that can be expressed as the sum of two___________________





Hardy: I rode in a taxi with the number 1729. An ordinary number

Ramanujan: No Hardy! It is a special number. It is the smallest number that can be exppressed as.............................

Fill in the blanks!

( Don't search the internet or run to the libray/bookstore for books on Ramanujan! :) )

[Benny Lava]

More hints:

Take the 3000 meter circular race track, place one sensor at 0 meters( or 3000 meters if you go around) and another sensor at 2000 meters.

For a 1000 meter race start at the first sensor and end at the second sensor.

For a 2000 meter race start at the second sensor(at 1000 meters) and finish at the first sensor(at 3000 meters or 0 meters).

For a 3000 meter race start at the first sensor and finish at the same sensor going around.

You need only two sensors and you do not have to change directions. But you moved the starting point. That is the key - moving the starting point for different lengths!

Hope this helps in the 31000 meters race course problem! :)

Yes. For a 31000 meter track, we have 30 different race distances to measure. If a combination of two sensors can measure for distance of N thousand meters, it can also be used to measure for distance of (31 - N) thousand meters. So essentially we need to have 15 combinations. 6C2 is equal to 15 hence 6 sensors are sufficient. Now the only problem is to decide how to place the 6 sensors such that all course lengths are covered :)

[Benny Lava]

Some more words of Ramanujan as hint:

Ramanujan: No Hardy! It is the smallest number that can be expressed as the sum of two___________________





Hardy: I rode in a taxi with the number 1729. An ordinary number

Ramanujan: No Hardy! It is a special number. It is the smallest number that can be exppressed as.............................

Fill in the blanks!

( Don't search the internet or run to the libray/bookstore for books on Ramanujan! :) )

I have heard this story before, but waiting for someone to answer :)

[GP]

naanum andha story kaelvi pattirukaen.
anyway one more specialty of the number 1729
1+7+2+9=19
multiply 19 by its reverse digit
i.e. 19 X 91 = 1729

[sathya_1979]

sum of 2 different sets of cubes:
10 power 3 + 9 power 3 = 1000 + 729 = 1729
12 power 3 + 1 power 3 = 1728 + 1 = 1729

[rajraj]

sum of 2 different sets of cubes:
10 power 3 + 9 power 3 = 1000 + 729 = 1729
12 power 3 + 1 power 3 = 1728 + 1 = 1729

Correct! :) 1729 is the smallest number that can be expressed as the sum of two cubes in two different ways !