   SEARCH HOME Math Central Quandaries & Queries  Question from Thomas, a student: Hi, i have got this brainteaser from a teacher who wants me to solve it, but i have no idea to proof my solution: You have the digits 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, 7, 7, 8, 8, 9, 9 in a random order side by side and put randomly between two digits a colon, so that a division comes into existence. Can the result of the calculation be 2? Thx in advance Thomas,

When solving any mathematics problem, one must state it in such a way that all the details are clear. Here is my interpretation of your question:

You are to distribute the given numbers into two groups with the product of the numbers in the first group equal to K and the product in the second group equal to L. The question asks if you can arrange the groups so that K = 2*L. If that is indeed the question, then the answer is pretty easy. The number of 2's in the factorization of one group must be even, the number of 2's in the other factorization must be odd.

Chris

Thomas replied

Hi again,

thank you for the first answer.
I'm sorry that my question was not as exact as necessary.

You are right that there is a term K = 2*L, but K and L are not equal to the products of the numbers, they just consist of the numbers, e.g. L = 23456789 and K would be 46913578 (in this example the condition that there are no 1's is not kept, of course) I am pretty sure that there is no possibility to find L and K in a way that all conditions are kept, but I need - if it isnt too much trouble for you - an precise and exact mathematcial proof.
I am looking forward to hearing from you

Nice!

Suppose that you have numbers A and B such that A/B = 2.

What is A+B in terms of B?

Is this possible? Hint: how do you test a long number to see if it is divisible by 3? How does this generalize to two big numbers that have a sum divisible by 3?

Good Hunting!
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