Question

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

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

Hint:

Repeatedly adding the digits keeps the remainder on division by 9 unchanged. Which numbers have seed 9?

Answer 1

The Correct Option is B

Step 1: Use the digit-sum rule for 9:
A number and the sum of its digits leave the same remainder when divided by 9. So repeating the digit sum does not change the remainder mod 9, and \(\text{seed}(n)\) is a single digit with the same remainder as \(n\).

Step 2: When is the seed 9?
For a positive integer, the digit sum is never 0, so the seed is a digit from 1 to 9. The seed equals 9 exactly when the remainder is 0, that is, when \(n\) is a multiple of 9.

Step 3: Count the multiples of 9 below 500:
The multiples are \(9, 18, \ldots, 495\). Since \(9 \times 55 = 495 < 500\) while \(9 \times 56 = 504 > 500\), there are \(55\) of them.

Final Answer:
The correct option is (2): 55.