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
Answer
14
Solution
The first digit can be chosen in 9 ways (1 to 9) and the 2nd to 6th digits in 105 ways.
For any choice of the first six digits, exactly 5 of the 10 possible last digits make the sum even, so the
required number = 9 105 5 = 45 105
m n 10n = 9 5 105
which gives n = 5, m = 9 (the only choice with m, n \1, , 9\).
m + n = 14
Answer: 14