The number of 3-digit odd numbers, whose sum of digits is a multiple of 7, is _____________.

Solution

Maths sir · 3 steps

Fix the odd units digit first, then count the pairs of first two digits that reach 7, 14 or 21.

1. Possible sums. Write the number as abc with a from 1 to 9, b from 0 to 9 and c ∈ {1, 3, 5, 7, 9}. The digit sum is at most 27, so it must be 7, 14 or 21.
2. Count digit pairs. Once c is fixed, you need a + b = t, where t is the target sum minus c. Because a ≥ 1, the number of pairs (a, b) is:

   N(t) = t, 1 ≤ t ≤ 9 ; 19 - t, 10 ≤ t ≤ 18 ; 0, otherwise

   Units digit c	Sum 7	Sum 14	Sum 21	Total
   1	a+b=6: 6	a+b=13: 6	a+b=20: 0	12
   3	a+b=4: 4	a+b=11: 8	a+b=18: 1	13
   5	a+b=2: 2	a+b=9: 9	a+b=16: 3	14
   7	a+b=0: 0	a+b=7: 7	a+b=14: 5	12
   9	a+b=-2: 0	a+b=5: 5	a+b=12: 7	12

3. So. Adding the totals gives 12 + 13 + 14 + 12 + 12 = 63, so there are 63 such numbers.

The common slip: 47. Forgetting that the digit sum can reach 21 leaves out 16 numbers and gives 47.

The idea to keep. When a digit is restricted, as the odd units digit is here, fix it first. Then count the remaining pairs with a + b = t, and remember the first digit cannot be 0.

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