Two-digit Sum And Product
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?
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?