If the number of seven-digit numbers, such that the sum of their digits is even, is m · n · 10ⁿ ; m, n ∈{1,2,3, …, 9}, then m+n is equal to__________

Solution

Maths sir · 4 steps

Fill the first six digits freely, then let the last digit fix the parity: exactly 5 choices work.

1. First six digits. The first digit cannot be 0, so it has 9 choices. Each of the next five digits has 10 choices. That gives 9 × 10⁵ ways.
2. Last digit. Whatever the first six digits add up to, the last digit must have a matching parity. If that sum is even, the last digit must be one of 0, 2, 4, 6, 8. If it is odd, the last digit must be one of 1, 3, 5, 7, 9. Either way there are exactly 5 choices.

   9 × 10⁵ × 5 = 9 · 5 · 10⁵

3. Match the form. Compare with m · n · 10ⁿ. The power of 10 gives n = 5, and then m · 5 = 45 gives m = 9. Both lie in {1, 2, …, 9}.
4. So. m + n = 9 + 5 = 14.

The common slip: 45. Reporting the product mn = 45 instead of the sum gives 45, not 14.

The idea to keep. For an even or odd digit sum, let all digits but one vary freely. The last digit then has exactly 5 valid choices, so the answer is half of the total count.

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