The largest product

Put the digits 2, 3 and 5 into a two-digit number times a one-digit number, using each digit once. Which arrangement gives the largest product?

Be systematic, not random. The first box can hold 2, 3 or 5, and for each choice the other two digits can go two ways. That gives six arrangements: 23 × 5, 25 × 3, 32 × 5, 35 × 2, 52 × 3, 53 × 2.

You could just multiply all six. But grouping is smarter. Pair them by the multiplier (the one-digit number): (35 × 2, 53 × 2), (25 × 3, 52 × 3), (23 × 5, 32 × 5). In each pair the bigger two-digit number clearly wins, so only 53 × 2, 52 × 3 and 32 × 5 are left.

53 × 2 loses to 52 × 3: the two-digit numbers are nearly the same, but 3 is a bigger multiplier. That leaves 52 × 3 and 32 × 5.

Split them into parts:

• 32 × 5 = (3 × 10 × 5) + (2 × 5) = 150 + 10.
• 52 × 3 = (5 × 10 × 3) + (2 × 3) = 150 + 6.

The first parts are the same, so the second part decides: 32 × 5 = 160 is the largest.

So the largest digit became the multiplier, and the other two went in decreasing order. Is that always true? Algebra decides.

Let the digits be p, q and r, with p < q < r. Grouping by multiplier and keeping the bigger two-digit number each time leaves rq × p, rp × q and qp × r. Since q > p, rp × q beats rq × p.

Finally split the last two: qp × r = (10 × q × r) + (p × r), and rp × q = (10 × r × q) + (p × q). The first parts are the same, and since r > q, the first one wins. So qp × r is always the largest: the largest digit as the multiplier, the other two in decreasing order.