Question

The seed of any positive integer \(n\) is defined as: \[ \text{seed}(n) = n, \ \text{if } n<10 \] \[ \text{seed}(n) = \text{seed}(s(n)), \ \text{otherwise} \] where \(s(n)\) is the sum of digits of \(n\). How many positive integers \(n\), such that \(n<500\), will have \(\text{seed}(n) = 9\)?

• A. 39
• B. 72
• C. 81
• D. 108
• E. 55

Hint:

Seed(n) = 9 corresponds to numbers divisible by 9; use divisibility rules to count them quickly.

Answer 1

The Correct Option is B

Numbers with seed 9 are those divisible by 9. We need to count multiples of 9 less than 500. Largest multiple: \(9 \times 55 = 495\). Number of such multiples: \(55\). But seed 9 also includes numbers like \(n = 9k\) where seed(\(n\)) reduces to 9. For \(n<500\), the number of integers divisible by 9 is: \[ \left\lfloor \frac{499}{9} \right\rfloor = 55 \] Hence answer is \(\boxed{55}\).