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**ganesh****Moderator**- Registered: 2005-06-28
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You are correct, wcy.

Good, you didn't tell how you did it

Try earlier problems too.

It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 22,096

Problem # k + 9

A square, whose side is 2 meters, has its corners cut away so as

to form an octagon with all sides equal. Then, what is the length of each side of the octagon, in meters?

It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,634

Ganesh, I am thinking of an upgrade to the "hide" tag.

Wouldn't it be cool if we could **really** hide the results from everyone (except moderators) until a certain time, maybe 1 week after the post, or based on the time you post the question.

A third possibility would be for you to have a keyword in the topic like "open" and that could be what controls whether the hide tag really hides, so "Ganesh's Puzzles" would be a list of solved/unsolved puzzles, with other peoples answers hidden until you let them be shown.

I would obviously need to program all this, and I am not sure the hurdles I would face, but what do you think of the idea?

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

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**ganesh****Moderator**- Registered: 2005-06-28
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The idea seems great!

Inspite of the hide tag, its too tempting for many to view the answer.

Since most of our members solve the problems and post their solutions in a day or two, I post a problem or two everyday. Once the problem is solved, I forget to tell how it is done.

Your idea would be of great help

It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**MathsIsFun****Administrator**- Registered: 2005-01-21
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So, do you think it would be good to post a new topic for every puzzle? That way they could have titles like "Cut a Square (Current)", which you can edit to "Cut a Square" when it is solved. I could use the phrase "(Current)" to tell the forum to **really** hide answers in the hide tag.

Also, if someone posts a full answer you can edit their post, insert a "hide" tag and put a gentle reminder to use the hide tag

(This is all based on the assumption I can get the hide tag to work like we want!)

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 22,096

Yes, that seems a good suggestion.

Two **tricky** problems to start the day:-

Problem # k + 10

Complete the series:-

1248, 1632, 6412, 8256, _____

Problem # k + 11

What is special about the number 2592?

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 22,096

Problem # k + 12

In a 4-digit number, the sum of the first two digits is equal to that

of the last two digits. The sum of the first and last digits is equal to the third digit. Finally, the sum of the second and fourth digits is twice the sum of the other two digits. What is the third digit of the number?

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**kylekatarn****Member**- Registered: 2005-07-24
- Posts: 445

solution(#k+9)

*Last edited by kylekatarn (2005-09-02 00:40:05)*

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**ganesh****Moderator**- Registered: 2005-06-28
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Solution to problem # k + 9

The side of the square is 2.

Let a length 'c' be cut from the two ends of a side.

We have right angled triangles of sides 'c'.

The hypotenuse would be h² = 2c²

or h = (√2)c

Since it is a regular octagon.

2 - 2c = (√2)c

c = 2/(2+√2)

h is the side of the octagon,

h = 2 (√2)/(2+√2)

or h = 2/(1+√2)

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**ganesh****Moderator**- Registered: 2005-06-28
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Problem # k + 13

Two squares are chosen at random on a chessboard. What is the probability that they have a side in common?

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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,588

I guess 1/18 chance if they can't both be the same chess square.

**igloo** **myrtilles** **fourmis**

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**ganesh****Moderator**- Registered: 2005-06-28
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You got it right! Well done! Try the earlier unsolved problmes.

The number of ways of choosing the first square is 64. The number of ways of choosing the second square is 63. There are a total of 64 * 63 = 4032 ways of choosing two squares.

If the first square happens to be any of the four corner ones, the second square can be chosen in 2 ways. If the first square happens to be any of the 24 squares on the side of the chess board, the second square can be chosen in 3 ways. If the first square happens to be any of the 36 remaining squares, the second square can be chosen in 4 ways. Hence the desired number of combinations = (4 * 2) + (24 * 3) + (36 * 4) = 224. Therefore, the required probability = 224/4032 = 1/18

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**ganesh****Moderator**- Registered: 2005-06-28
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Problem # k + 14

What is the area of the largest triangle that can be fitted into a rectangle of length 'l' units and width 'w' units?

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**John E. Franklin****Member**- Registered: 2005-08-29
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*Last edited by John E. Franklin (2005-09-04 14:05:54)*

**igloo** **myrtilles** **fourmis**

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**John E. Franklin****Member**- Registered: 2005-08-29
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*Last edited by John E. Franklin (2005-09-04 14:42:33)*

**igloo** **myrtilles** **fourmis**

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**John E. Franklin****Member**- Registered: 2005-08-29
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**igloo** **myrtilles** **fourmis**

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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,588

**igloo** **myrtilles** **fourmis**

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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,588

**igloo** **myrtilles** **fourmis**

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 22,096

Good work, John E. Franklin

Solution to Problem # k + 14

Any triangle you try to draw with the maximum area would have same base and height. The solution is, without doubt, lw/2. You can try all possibilities. You'd get the same answer, both when the base is 'l' and 'w'.

Solution to Problem # k + 12

Yes, you would be one equation short. But, when you get one number is eight times the other, the numbers would have to be 1 and 8, as each number a,b,c, and d is a single digit number!

Solution to Problem # k + 10

Yes, 5121 is the correct answer. As you said, there may not be another as 0242 is not acceptable!

Solution to Problem # k + 11

You have given a different property! I had this is my mind.

It is a number of the form abcd equal to (a^b)*(c^d)

Solution to Problem # k + 6

You are correct. The ages of the grandmother and the grandson would be (61,1), (62,2), (63,3), (64,4), (65,5), and (66,6).

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**ganesh****Moderator**- Registered: 2005-06-28
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Problem # k + 15

A sample of x litres from a container having a 60 litre mixture of milk and water containing milk and water in the ratio of 2 : 3 is replaced with pure milk so that the container will have milk and water in equal proportions. What is the value of x?

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**John E. Franklin****Member**- Registered: 2005-08-29
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**igloo** **myrtilles** **fourmis**

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**John E. Franklin****Member**- Registered: 2005-08-29
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Thanks ganesh. Yeah, I see now on #k + 12 that you can solve that

"b" is twice "d", and "d" is 4 times bigger than "a".

That's really something.

**igloo** **myrtilles** **fourmis**

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**ganesh****Moderator**- Registered: 2005-06-28
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Solution to Problem # k + 6

You are right, John. I don't know the reason. Just as MathsIsFun thought, I too believed 97531 x 86420 would be the highest product!

Problem # k + 16

The pages in a book are serially numbered from 1. If the number of digits required to total all the pages in the book is 972, how many pages are there in the book?

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**John E. Franklin****Member**- Registered: 2005-08-29
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**igloo** **myrtilles** **fourmis**

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**ganesh****Moderator**- Registered: 2005-06-28
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John, you went wrong somewhere; please try again

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