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Digit Sum Seed Function

CAT 2008 Slot 1 · Quantitative Ability · Medium · Number System

This is a medium Quantitative Ability question from the CAT 2008 Slot 1 paper. It tests Number System. The full answer key and a step-by-step explanation are below — try it yourself first, then reveal the solution.

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

  1. A.

    39

  2. B.

    72

  3. C.

    81

  4. D.

    108

  5. E.

    55

Answer

E

Explanation

The seed(n)\text{seed}(n) function gives the digital root (remainder when nn is divided by 9, with remainder 0 represented as 9).

Thus, seed(n)=9\text{seed}(n) = 9 for all multiples of 9.

We need to find the number of positive integers n<500n < 500 that are divisible by 9. ⌊4999⌋=55\left\lfloor \frac{499}{9} \right\rfloor = 55.

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