Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
What have you tried? –  vadim123 May 7 '13 at 12:51
2  
Obviously I tried and get the answer $56^3+56*25^2$. –  ᴊ ᴀ s ᴏ ɴ May 7 '13 at 12:53
1  
We can help you better if you share the details of your solution. –  vadim123 May 7 '13 at 12:57
add comment

1 Answer

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.

share|improve this answer
    
AH HA! Thanks!! –  ᴊ ᴀ s ᴏ ɴ May 7 '13 at 13:05
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.