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March 15th, 2010, 08:10 AM   #1
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8th grade proble, challenging???

Problem 1

On each face of a cube a certain natural number was written, and at each vertex a number equal to the product of the numbers on the three faces adjacent to that vertex was placed. If the sum of the numbers on the vertexes is 70 then what is the sum of the numbers on all the faces of the cube?


a, 12 b, 35 c, 14 d, 10 e, cannot be determined


Problem 2

Pages problem?
math grade7
Peter put magazines on a bookshelf. They have either 48 or 52 pages. which of the following number cannot be the total pages of all the magazines on the bookshelf?

a, 500 b, 524 c, 568 d, 588 e, 620
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March 15th, 2010, 09:34 AM   #2
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Re: 8th grade proble, challenging???

Quote:
Originally Posted by Recycle98
Problem 1

On each face of a cube a certain natural number was written, and at each vertex a number equal to the product of the numbers on the three faces adjacent to that vertex was placed. If the sum of the numbers on the vertexes is 70 then what is the sum of the numbers on all the faces of the cube?


a, 12 b, 35 c, 14 d, 10 e, cannot be determined
I have no idea how an 8th grader could prove the answer to this question. My best guess is that they should find a solution and see if it's on the list, guessing that it's the only one of they can find no others. It's clear enough that it cannot be 35, since the sum would have to be at least 30*1*1 + 30*1*1 + 30*1*1 + 30*1*1 + 1*1*1 + 1*1*1 + 1*1*1 + 1*1*1 = 124.

The correct answer is 14.

Quote:
Originally Posted by Recycle98
Problem 2

Pages problem?
math grade7
Peter put magazines on a bookshelf. They have either 48 or 52 pages. which of the following number cannot be the total pages of all the magazines on the bookshelf?

a, 500 b, 524 c, 568 d, 588 e, 620
Nice question. First, all page numbers must be multiples of 4 since 48 and 52 are -- though in this case all answers are also, so no easy answer there. The key to the solution is seeing that you have a certain number of magazines that account for the bulk, then adjust as needed with the balance to get the right total. For 500, start with rounddown(500/4 = 10 magazines of 48 pages, which gives 480 pages total. You can see that changing 5 magazines to longer ones makes the total 480 + 5*4 = 500. Do the same for the others.

http://en.wikipedia.org/wiki/Coin_problem also gives the solution, but it's probably harder to use that more powerful technique here.
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March 15th, 2010, 01:35 PM   #3
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Re: 8th grade proble, challenging???

Problem 1:

Labeling the faces a, b, c, d, e and f:

abe + bce + cde + ade + abf + bcf + cdf + adf = 70

(e + f)(ab + bc + cd + ad) = 70

(a + c)(b + d)(e + f) = 70. Factors of 70: 2, 5, 7, 10, 14, 35.
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March 15th, 2010, 03:17 PM   #4
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Re: 8th grade proble, challenging???

Nice, Greg.
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March 15th, 2010, 03:18 PM   #5
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Re: 8th grade proble, challenging???

ty
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