(a)+(b)+(c)+(d)+(e) odd numbers

There are 5 questions in a quiz designated by:

a, b, c, d, e

Each of the questions can only have one of 4 possible answers:

Possible Answer #1: 5 Possible Answer #2: -5 Possible Answer #3: 7 Possible Answer #4: -7

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Am I correct to say that there are 625 possible variations of answers that we can receive?

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Can any of the variations equal to zero when we would add all the answers up?

For example on variation

(5) + (-5) + (7) + (-7) + (5) = 5

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So the core question is can the sum of all the answers to a particular variation ever be zero.

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Do you mean 5,-5,7, and -7? Also, can you explain the last sentence a little more clearly? –  user3180 Feb 4 '11 at 7:57

2 Answers

The answer is no. The sum of a single odd number is odd (since it is just the original number), the sum of two odd numbers is even (see here), and in general, the sum of $n$ odd numbers is odd if $n$ is odd, and even if $n$ is even.

In your problem, you are adding 5 odd numbers, and because 5 is odd, we must have that the sum is odd, regardless of which of the values $5,-5,7,-7$ the numbers $a,b,c,d,e$ take on. However, 0 is an even number. Thus, the sum $a+b+c+d+e$ can never be 0.

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As to the first question ("Am I correct to say that there are 625 possible variations of answers that we can receive?"), if I understand correctly the setting, you have four possible answers for the first question, four for the second and so on. So the number of possible sets of five answers is 4 × 4 × 4 × 4 × 4, i.e. 4^5, i.e 1024.

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