Question

Suppose, the seed of any positive integer $n$ is defined as follows: $$	ext{seed}(n) = n, 	ext{ if } n < 10$$ $$= 	ext{seed}(s(n)), 	ext{ otherwise,}$$ where $s(n)$ indicates the sum of digits of $n$. For example, $	ext{seed}(7) = 7$, $	ext{seed}(248) = 	ext{seed}(2 + 4 + 8) = 	ext{seed}(14) = 	ext{seed}(1 + 4) = 	ext{seed}(5) = 5$ etc. How many positive integers $n$, such that $n < 500$, will have $	ext{seed}(n) = 9$?

Accepted Answer

E. The $	ext{seed}(n)$ function gives the digital root (remainder when $n$ is divided by 9, with remainder 0 represented as 9). Thus, $	ext{seed}(n) = 9$ for all multiples of 9. We need to find the number of positive integers $n < 500$ that are divisible by 9. $$\left\lfloor \frac{499}{9} ightfloor = 55$$.

Digit Sum Seed Function

Explanation

Practise this under exam conditions