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**Devantè****Real Member**- Registered: 2006-07-14
- Posts: 6,400

3e) 999 + 9^0(9^0)(9^0)(9^0)(9^0)(9^0)

I'll see if I can find another way.

*Last edited by Devanté (2006-08-21 18:52:52)*

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**tt****Member**- Registered: 2005-07-03
- Posts: 32

John E. Franklin wrote:

28) I get 44 times at right angles. I was too lazy, I only counted 22 from 6am to 6pm.

What is the answer for a seven-hour clock and a fourteen hour day?? Do you know Ganesh? I don't.

The reason I ask is because with a 12 hour clock the 9 and the 3 are at right angles to the 12 at the top, so

the answer is less than 12 x 2 for 12 hours, but a seven hour clock would be weird, or a 5 hour clock too.

Hmmm.... I'll have to ponder this more later...

And I only see 42 times accounted for! I must have lost count around the dial.

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**tt****Member**- Registered: 2005-07-03
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*Last edited by tt (2006-09-03 22:44:04)*

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**tt****Member**- Registered: 2005-07-03
- Posts: 32

ganesh wrote:

32. A, B, C, D, E, and F are sitting on six chairs placed around a round table, not necessarily in that order. A is between D and F. C is opposite D and D and E are not sitting next to each other. Which of these pairs are sitting next to each other?

(i) A and B (ii) C and E (iii) B and F (iv) A and C

My answer

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**ganesh****Moderator**- Registered: 2005-06-28
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48. What is the maximum value of n such that 100! (100 factorial) is divisible by 6^n?

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**mathsyperson****Moderator**- Registered: 2005-06-22
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Why did the vector cross the road?

It wanted to be normal.

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**ganesh****Moderator**- Registered: 2005-06-28
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Absolutely right, mathsyperson! Well done!!!

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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**landof+****Member**- Registered: 2007-03-24
- Posts: 131

oh my! I cannot believe no one answered (10)!!! It is so easy!

*Last edited by landof+ (2007-03-30 00:03:41)*

I shall be on leave until I say so...

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

28.)

tt wrote:

And I only see 42 times accounted for! I must have lost count around the dial.

Well, I don't blame you, I just counted in my head, and only got 20 once around.

But then 10 minutes later I tried again and got 22 again. So it's easy to mess up in your head.

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

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**Eeyore****Member**- Registered: 2007-04-16
- Posts: 8

Maybe I just missed it, but I never did see an answer to #5.

I don't really see anything that would exclude the possibility that the two numbers are the same number, so... 0 and 0 would be the smallest.

If you insist that these numbers are not only arbitrary whole numbers, but also disctinct, then I guess 1 and 0 would be the smallest.

Am I understanding the question here??

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**Eeyore****Member**- Registered: 2007-04-16
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the answer to #4:

she sold 3 to the first customer, 1.5 to the second customer and .5 to the 3rd customer for a total of 5.

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**mathsyperson****Moderator**- Registered: 2005-06-22
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I would guess that question 5 wants you to find two distinct natural numbers, as allowing them to be equal or 0 would trivialise the question.

The answer to 4 is slightly wrong as well. The 3rd customer doesn't get the right amount of oranges, and the question also states that she never cuts an orange in half. You're nearly there though, I think you just made a slight mistake there and it carried forward.

Why did the vector cross the road?

It wanted to be normal.

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**landof+****Member**- Registered: 2007-03-24
- Posts: 131

4. The total I got is 7. Eeyore, you must have made some calculation errors.

The girl sold half of what she had and 1/2 of an apple. The 3rd customer got (0+1/2)*2, which is 1. the second got:

1+1/2*2=3

3-1=**2**

The third got:

(3+1/2)*2=7

7-3=**4**

The total would be 4+2+1=7.

I shall be on leave until I say so...

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**Eeyore****Member**- Registered: 2007-04-16
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good job Landof, I don't know what I was thinkin, if she started with 3 and did all the stuff the problem says she did, she would still have half an orange left.

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**ganesh****Moderator**- Registered: 2005-06-28
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49. How many squares are there in all in a chess board? (Treating 2squares by 2 squares as a square etc.)

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**Identity****Member**- Registered: 2007-04-18
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*Last edited by Identity (2007-05-30 20:11:46)*

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**JaneFairfax****Member**- Registered: 2007-02-23
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Using the same method as above, you should also be able to prove that the total number of rectangles on a chessboard (counting squares as rectangles) is

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**ganesh****Moderator**- Registered: 2005-06-28
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Outstanding, Identity and Jane Fairfax, for that simple and elegant proof!!!

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**ganesh****Moderator**- Registered: 2005-06-28
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50A. A farmer has a piece of land of the shape given in the uploaded image (1001A). He has four sons. He desires divinding the land amongst his four sons. The four sons aren't very smart, they insist on the pieces of land being of the same area as the original land and also the same shape.

How would he divide them as closely adhering to their conditions as possible?

50B. Problem Same as 50 A. However, the land owned by the farmer is this time of the shape given in uploaded image 1002A. Same concundrum, same condition. How did he do it?

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**TheDude****Member**- Registered: 2007-10-23
- Posts: 361

*Last edited by TheDude (2007-11-12 04:35:19)*

Wrap it in bacon

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**ganesh****Moderator**- Registered: 2005-06-28
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Solution to #50!

TheDude,

Yours ASCII Art is good, you have drawn the picture neatly!

Plus, your answer is perfectly correct!!!**Well done, TheDude! Outstanding!**

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**ganesh****Moderator**- Registered: 2005-06-28
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51. There are two candles. Both are of same height but different diameters. One burns for 4 hours and the other burns for 3 hours. Both of them are lit at the same time. After what time would one be double the height of the other?

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**ganesh****Moderator**- Registered: 2005-06-28
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52. In a get-together, every person present shakes hands with every other person once. If there are 91 handshakes in all, how many people attended the get-together?

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**NullRoot****Member**- Registered: 2007-11-19
- Posts: 162

Trillian: Five to one against and falling. Four to one against and falling Three to one, two, one. Probability factor of one to one. We have normality. I repeat, we have normality. Anything you still cant cope with is therefore your own problem.

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**ganesh****Moderator**- Registered: 2005-06-28
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**Excellent, NullRoot!!! Your answer is perfectly right!!!**

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