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# Ahmes's Papyrus

### #11

Posted 28 November 2007 - 11:04 AM

rookie1ja (site admin)

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### #12

Posted 30 November 2007 - 05:16 PM

A represents worker 5

X represents the constant distance between the amount of corn that each worker has more than the last.

A + (A-X) + (A-2X) + (A-3x) + (A-4x) = 100

7(A-3X + A-4X) = A + (A-X) + (A-2X)

Some of this has some similarities to the above solution obviously. This is actually the extended expression of what is in the solution above. I simplified it in a very different fashion, which wasn't near as simple, perhaps.

If you solve for X in the second expression your end result is as follows:

A = X*46/11

With that in mind, I replaced all instances of A with X*46/11 and solved for X.

X = 9.16667 (55/6)

A = 55/6 * 46/11 OR 5/6 * 46 = 38.3333...

Obviously from there, you can deduce the rest and the results are identical. I have a feeling that I did this the hard way.

### #13

Posted 06 December 2007 - 09:23 PM

Here's a site that talks about Egyptian bread:

http://www.touregypt.net/featurestories/bread.htm

### #14

Posted 10 January 2008 - 07:23 AM

2nd worker = 10.83333 measures of corn

3rd worker = 20 measures of corn

4th worker = 29.16666 measures of corn

5th worker = 38.3333 measures of corn

**Edited by miya, 10 January 2008 - 07:25 AM.**

### #15

Posted 21 January 2008 - 07:43 AM

2 worker : 10 5/6 m

3 worker : 20m

4 worker : 32 1/2m

5 worker : 238 1/3m

### #16

Posted 15 February 2008 - 04:21 PM

Ahmes's Papyrus- Back to the Logic Puzzles

About 2000 B. C. there lived Ahmes, a royal secretary and mathematician of the Pharaoh Amenemhat III. In 1853 an Englishman Rhind found one of Ahmes's papyruses near the temple of Ramses II. in Thebes. The papyrus is a rectangle 33 cm wide and about 5 m long. There is the following math brain teaser on it (besides others).

100 measures of corn must be divided among 5 workers, so that the second worker gets as many measures more than the first worker, as the third gets more than the second, the fourth more than the third and the fifth more than the fourth. The first two workers shall get seven times less measures of corn than the three others.

How many measures of corn shall each worker get?

Edit: you can have fractional measures of corn.Spoiler for Solution

I suck at math but I love Logic problems. I can't stand fractions so my answer may be worng as ever but this seems quite simple. If you split up 100 measures of corn amongst 5 poeple and the first person has to have less than the second an so on but also have 7 times less than the 5th person than wouldn't the outcome look like this:

1st person- 5 measures of corn

2nd person- 10 measures

3rd- 20 measures

4th- 30 measures

5th 35 measures.

That all adds up to 100 and 35 divided by 5 is 7, right? Which makes the first person having 7x less than the 5th.

### #17

Posted 15 February 2008 - 05:28 PM

that does not work because there are various differences between the workers (1st vs. 2nd = 5 measures, but 2nd vs. 3rd = 10 measures ...)I suck at math but I love Logic problems. I can't stand fractions so my answer may be worng as ever but this seems quite simple. If you split up 100 measures of corn amongst 5 poeple and the first person has to have less than the second an so on but also have 7 times less than the 5th person than wouldn't the outcome look like this:

1st person- 5 measures of corn

2nd person- 10 measures

3rd- 20 measures

4th- 30 measures

5th 35 measures.

That all adds up to 100 and 35 divided by 5 is 7, right? Which makes the first person having 7x less than the 5th.

rookie1ja (site admin)

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### #18

Posted 16 February 2008 - 12:52 AM

### #19

Posted 13 March 2008 - 05:37 PM

Ahmes's Papyrus- Back to the Logic Puzzles

About 2000 B. C. there lived Ahmes, a royal secretary and mathematician of the Pharaoh Amenemhat III. In 1853 an Englishman Rhind found one of Ahmes's papyruses near the temple of Ramses II. in Thebes. The papyrus is a rectangle 33 cm wide and about 5 m long. There is the following math brain teaser on it (besides others).

100 measures of corn must be divided among 5 workers, so that the second worker gets as many measures more than the first worker, as the third gets more than the second, the fourth more than the third and the fifth more than the fourth. The first two workers shall get seven times less measures of corn than the three others.

How many measures of corn shall each worker get?

Edit: you can have fractional measures of corn.Spoiler for Solution

If the 7 were changed to 3, ie. "The first two workers shall get THREE times fewer measures of corn than the three others.", then the consequence will be an integer solution requiring no corn-chopping. NINE works too, but leaves the first worker without corn -- not fair :-(

### #20

Posted 20 March 2008 - 03:45 PM

If the 7 were changed to 3, ie. "The first two workers shall get THREE times fewer measures of corn than the three others.", then the consequence will be an integer solution requiring no corn-chopping. NINE works too, but leaves the first worker without corn -- not fair :-(

Nice work belwood. I like the idea of making it an integer solution. Even though cleaning it up makes it seem easier it's still a fun puzzle.

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