Question

Let s(n) and p(n) denote the sum of all digits of n and
the product of all digits of n (when written in decimal
form), respectively. S is the sum of all two-digit natural
numbers n such that n = s(n) + p(n). Find the sum of
digits of s.​

Answer 1

Answer:

9

Step-by-step explanation:

Let the no. be 10a + b

Given that 10a + b = a + b + ab

    10a = a(1+b)

b + 1 = 10  => b = 9

So, the no. can be anything but should end with 9, as we see that the tens digit does not matter.

So, the possible values = 19, 29, 39, 49, 59, 69, 79, 89, 99.

So the sum of no.s  = 19+29+39+49+59+69+79+89+99 = 531 = S

So, sum of digits of S = 5+3+1 = 9