*Spoiler for New Scientist Enigma 1727: “Common Factors” *(Follow the link to see the puzzle.)

If *b* is the middle number then the sum of the three is 3*b* and the product is *b*(*b*^{2}–1). For 3*b* to have at least six factors, *b* must be composite.

Clearly all the factors of *b* are common factors of both the sum and the product. So are 3 times those factors, if (*b*^{2}–1) is divisible by 3. But one of *b*–1, *b*, *b*+1 is divisible by 3, and if it’s not *b* then (*b*^{2}–1) is divisible by 3. So if *b* is not divisible by 3, and has *n* factors, then both 3*b* and *b*(*b*^{2}–1) have at least 2*n* factors in common. Since *b* is composite it has three distinct factors if it is the square of a prime and more than three distinct factors otherwise. For the number of common factors to be 6, then, *b* must be the square of a prime if it is not divisible by 3.

But if *b* is the square of a prime (and is a two-figure number, therefore not 4) then it is odd, its factors are odd, and all six factors of 3*b* are odd. Their sum is even.

So *b* instead must be divisible by 3.

In that case all the factors of *b* are common factors of 3*b* and *b*(*b*^{2}–1), but there are no others. So *b* must have exactly six distinct factors. For that to happen *b* must be of the form *pq*^{2}, where *p* and *q* are both prime. The factors of *b* are 1, *p*, *q*, * q^{2}*,

*pq*, and

*pq*

^{2}.

If *p* and *q* are both odd then all six factors are odd and their sum is even. If *p* is odd and *q* is even then again the sum is even. If *p* is even and *q* is odd then the sum is odd. So *q* is 3 and *p* is even, and there aren’t many even primes.

*b* =* pq*^{2 }= 18. 17+18+19 = 54 with factors 1, 2, 3, 6, 9, 18 summing to 39; 17 x 18 x 19 = 5814 having those six factors among others.