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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
We can help you better if you share the details of your solution. – vadim123 May 7 '13 at 12:57
up vote 5 down vote accepted

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.

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AH HA! Thanks!! – ᴊ ᴀ s ᴏ ɴ May 7 '13 at 13:05

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