My son was given the following problem in his math club.

We can see that 2 + 2 = 4 and 2 × 2 = 4 and similarly 1 + 2 + 3 = 6 and 1 × 2 × 3 = 6.

Can you find four whole numbers where their sum and their products are the same? What about five whole numbers?

Spoiler alert: If you read further, you will get the solution to this problem. You may want to try it out yourself first. Scroll down when you are ready.

My son found 1 + 1 + 2 + 4 = 8, 1 × 1 × 2 × 4 = 8 and 1 + 1 + 1 + 2 + 5 = 10, 1 × 1 × 1 × 2 × 5 = 10. From this he conjectured that if you take a bunch of 1s followed by 2 and then followed by the sum of the 1s and the 2, that this list of numbers has the property that it multiplies and adds to give twice the last number in the list.

I asked him if he could prove that his conjecture is always true. He said, “The number of numbers minus two is the number of ones we need. Then plus two, that’s now the same value as the number of numbers at the end. If you add all of those numbers, you’ll get twice the number. And if you multiply all of those numbers you get 2 times the number, same value as the sum, because none of the 1s changes the value.”

“Or,” he continued, “If the number of numbers is n, then the number of 1s is n – 2, followed by a 2, followed by n. n – 2 + 2 is n. So the sum is 2n. The product is 1 times 1 times … and so on times 1 and then times 2 and times n which is also 2n.”