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 \(\text{seed}(n) = 9\)?
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 \(\text{seed}(n) = 9\)?
Show Hint
Seed(n) = 9 corresponds to numbers divisible by 9; use divisibility rules to count them quickly.
Updated On: Mar 24, 2026
- 39
- 72
- 81
- 108
- 55
Show Solution
Verified By Collegedunia
The Correct Option is B
Solution and Explanation
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 \(\boxed{55}\).
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