# 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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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 $$d=6$$.