# Sunday Times Teaser 2839 – Unwholesome

*by Mike Fletcher*

#### Published 19 February 2017 (link)

I have written down three positive numbers, the highest of which is an even two-figure number. Also, one of the numbers is the average of the other two. I have calculated the product of the three numbers and the answer is a prime number.

You might be surprised that the product of three numbers is a prime but, of course, they are not all whole numbers — at least one of them is a fraction.

What is the largest of the three numbers, and what is the product of the three?

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There is a straightforward analytic solution to this one (a relief after last weeks effort!).

If we let the smallest number be the simple fraction \(n/d\) and the largest be the integer \(2x\), the product of the three numbers, which is a prime, is then given by: \[\frac{x n (2 d x+n)}{d^2} = p\] Since the fraction is in its lowest terms, \(d\) does not divide \(n\) and this means that it does not divide \(2 d x + n\) either. So \(d^2\) must divide \(x\), which gives \[\left(\frac{x}{d^2}\right)n(2 x d + n) = p\]Since the product is prime, two of its three integer terms must be unity; and, since \(2 x d + n\) cannot be unity, we must have \(n=1\) and \(x=d^2\), which gives: \[2d^3+1 = p\] Since \(2x\) (= \(2d^2\)) is a two digit integer, we know that \(2 < d < 8\) which quickly leads to solutions for \(d=5\) and \(d=6\).

So it seems that there is a flaw in this teaser, which has two possible answers (50, 251) or (72, 433).