# Find the quantity of such numbers

A six digits number $abcdef$ satisfy:

1. Each digits is non zero.

2. $ab+cd+ef$ is even.

Find the number of such six digits numbers.

The answer given is $56^3+3 \cdot 56 \cdot 25^2$, why we need to multiply that 3? Thank you.

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Obviously I tried and get the answer $56^3+56*25^2$. – ᴊ ᴀ s ᴏ ɴ May 7 '13 at 12:53
Hint: If $ab + cd + ef$ is even then either
• $ab,cd,ef$ are all even
• exactly 2 of $ab,cd,ef$ are odd and one is even.
In the second part, you have a choice which one should be even, and there are exactly 3 possibilities. That explains the $3$ in the answer.