Question

If the number of seven-digit numbers, such that the sum of their digits is even, is $ m \cdot n \cdot 10^a $; $ m, n \in \{1, 2, 3, ..., 9\} $, then $ m + n $ is equal to

Hint:

Half of the total 7 digit numbers will have an even sum of digits.

Answer 1

Correct Answer: 14

Total 7 digit numbers = 9000000 7 digit numbers having sum of digits even 
= 4500000 = \( 9.5 \cdot 10^5 \) 
\( m = 9 \), \( n = 5 \) \( m + n = 14 \)