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Prove this if you can!

If you ask someone to choose ten random digits (not including 0) and multiply them all together, then to tell you all but one of the digits of the answer, you can predict the remaining digit.

Good luck proving it!

Note by Victor Loh
2 years, 9 months ago

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What do you mean? If I choose ten '1's, their product is 1. So I will tell that person nothing (all but one). However there are many possible digits. Joel Tan · 2 years, 9 months ago

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If there were a dislike button I would have pressed it. King Zhang Zizhong · 2 years, 7 months ago

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Let the random digits be \(a_1, a_2, a_3, \ldots, a_{10}\) \(\Huge\mbox{NOT}\) respectively, where \(a_{10}\) is the digit not told to you.

The product will be \(P=a_1a_2a_3\ldots a_{10}\), and the digits told to you will be \(a_1, a_2, a_3, \ldots, a_9\).

The remaining digit is obtained by dividing \(P\) by \(a_1, a_2, a_3, \ldots, a_9\) one by one, which equals to \(a_{10}\). Kenny Lau · 2 years, 9 months ago

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@Kenny Lau This is an extremely fun question, as well as easy. :) Kenny Lau · 2 years, 9 months ago

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