mathschallenge.net

Problem

If you multiply together the digits of the number 42, 4 2 = 8, but if you add the digits together, 4 + 2 = 6. For the number 31, the product of the digits, 3 1 = 3 and the sum of the digits, 3 + 1 = 4.

Can you find a two-digit number for which the product of its digits is the same as the sum of its digits?

Solution

Given the 2-digit number. ($ab$), we are solving: $a$ + $b$ = $ab$.

Therefore $ab$ $b$ = $a$, $b$($a$ 1) = $a$, giving $b$ = $a$/($a$ 1).

Considering the possible values of the digit, $a$: $b$ = 2/1, 3/2, 4/3, ..., 9/8 and the only integer solution is 2/1; that is, $a$ = 2 and $b$ = 2.

Hence there is only one two-digit solution, 22.

For which 2-digit numbers do the product of their digits exceed their sum?
Can you find any 3-digit number for which the sum of the digits is equal to the product of its digits?
What about $n$-digit numbers?