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m12, q30 --wrong answer?

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m12, q30 --wrong answer? [#permalink] New post 02 Sep 2009, 22:10
The mean of four integers will not change if all the integers are multiplied by any constant. What is always true about this set of numbers?

I. The mean of the set is 0.
II. The sum of the largest member and the smallest member of the set is 0.
III. The set contains both positive and negative integers.

.
.
.
.
According to the exam, statement I. is correct with the explanation, "if the constant is 0, the mean of the new set is 0. Thus, the mean of the original set has to be 0 as well."

However, this seemingly does not seem to be true when both the set of 4 integers and the constant is 1. If all integers are 1, the mean is 1, and if all integers are multiplied by 1, the mean will still be 1. Can someone please help me understand my mistake? Thanks!
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Re: m12, q30 --wrong answer? [#permalink] New post 05 Sep 2009, 10:28
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Hi dpgxxx,

You seem to be having trouble with the word "any." When their asking for "any constant," they don't mean "any one", they mean "any possible" constant. Thus, it doesn't matter if a set of numbers happens to work when you multiply it by one, unless it ALSO works with 2, pi, or 5,000,000. Therefore, since we have to have the same mean when multiplied by zero (in addition to all those other big numbers), the average must be 0. II is not true because [-1, -1, -1, 3] is a valid set, and III isn't because os [0, 0, 0, 0,]
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Re: m12, q30 --wrong answer?   [#permalink] 05 Sep 2009, 10:28
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