The number of natural numbers, between 212 and 999, such that the sum of their digits is 15, is _______.

Solution

Maths sir · 4 steps

Fix the hundreds digit first; then just count the pairs of last two digits that make up the rest of 15.

1. Set up. Write the number as a̅b̅c̅ with a + b + c = 15 and 0 ≤ b, c ≤ 9. Then b + c = 15 - a. The number 999 has digit sum 27, so it never counts. That leaves only the lower limit 212 to check.
2. Count digit pairs. For b + c = s with both digits from 0 to 9, there are s + 1 pairs when s ≤ 9. There are 19 - s pairs when s ≥ 9, because digits above 9 are not allowed.

   Hundreds digit a	b + c	Number of pairs
   2	13	6 (from 249 up to 294, all above 212)
   3	12	7
   4	11	8
   5	10	9
   6	9	10
   7	8	9
   8	7	8
   9	6	7

3. Check the boundary. Numbers starting with 1 are below 212, so they are left out. For a = 2 the smallest number is 249, so all 6 of them lie above 212 and count.
4. So. Total = 6 + 7 + 8 + 9 + 10 + 9 + 8 + 7 = 64.

The common slip: 69. Counting all three-digit numbers with digit sum 15 adds the 5 numbers that start with 1 (159 to 195), which are below 212. That gives 69, not 64.

The idea to keep. For digit-sum counting, fix one digit and count pairs for the rest: s+1 pairs if s ≤ 9, 19-s if s ≥ 9. Then check the range limits separately.

Answer from the NTA final answer key. Solution by PrepWiser.