Enigma 1727

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 3b and the product is b(b2–1). For 3b 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 (b2–1) is divisible by 3. But one of b–1, bb+1 is divisible by 3, and if it’s not b then (b2–1) is divisible by 3. So if b is not divisible by 3, and has n factors, then both 3b and b(b2–1) have at least 2n 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 3b 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 3b and b(b2–1), but there are no others. So b must have exactly six distinct factors. For that to happen b must be of the form pq2, where p and q are both prime. The factors of b are 1, pqq2pq, and pq2.

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= 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.

 

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