jmc

Question

The numbers 2, 3, 5, 7, 11, 13 are arranged in a multiplication table, with three along the top and the other three down the left. The multiplication table is completed and the sum of the nine entries is tabulated. What is the largest possible sum of the nine entries?

\times d e f a b c

Accepted Answer

The sum of the nine entries is ad + bd + cd + ae + be + ce + af + bf + cf = (a + b + c)(d + e + f).Note that the sum (a + b + c) + (d + e + f) = 2 + 3 + 5 + 7 + 11 + 13 = 41 is fixed, so to maximize (a + b + c)(d + e + f), we want the two factors to be as close as possible, i.e. 20 21 = 420. We can achieve this by taking a,b,c = 2,5,13 and d,e,f = 3,7,11, so the maximum sum is 420.