If the number of seven-digit numbers such that the sum of their digits is even is m n 10^n…

Question

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

Answer: 14

Accepted Answer

By symmetry, half of all 7-digit numbers have an even digit sum. Total 7-digit numbers $= 9\times10^6$.

$$\text{Number with even digit sum} = \frac{1}{2}\cdot9\cdot10^6 = 4.5\times10^6 = 9\times5\times10^5$$

So $m=9,\ n=5$ (matching form $m\cdot n\cdot10^n$): $9\times5\times10^5 = 4.5\times10^6$

$m+n = 9+5 = 14$

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