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 d1 or more ? (Posted on 2016-03-07)
A problem from the 2004 Harvard-MIT Math Tournament:

Zach chooses five numbers from the set {1, 2, 3, 4, 5, 6, 7} and tells their product to Claudia. She finds that this is not enough information to tell whether the sum of Zachâ€™s numbers is even or odd. What is the product that Zach tells Claudia?

I believe it is a d1 puzzle.

Are you on my side?

 See The Solution Submitted by Ady TZIDON No Rating

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 thoughts | Comment 1 of 4
I'll make up a term: prone: a natural number that's either prime or is 1.

It's easer to consider the two numbers that are left out, as the product is 7! divided by the product of the two left-out numbers, and the total is the total of all seven (clearly even, containing four odd numbers) minus the sum of the two left-out numbers, so the five numbers have the same parity as the two that are left out.

The two that are left out can't both be prones, as that would tell Claudia what they are. One or both must be composite.

It could immediately come to one's attention that 12 = 3*4 = 6*2 would cause the situation described and is a solution to the problem. As it's assumed that the puzzle has a unique solution, we could say that this is the solution (or rather, leads to the solution; see the next paragraph). But does this need to be verified as unique?  And if so, how hard is that verification? There are only a few products involving 4 and/or 6.

Of course all this working with the two numbers not included, may make the solver forget that what was sought was the produce of the other five numbers, and incorrectly answer 12, as the product sought. The correct answer of course is 7!/12 = 420.

Do the insights needed and the possible proneness to error make it a D2? It's hard to say.

Edited on March 7, 2016, 9:51 am
 Posted by Charlie on 2016-03-07 09:49:56

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