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If a number is a multiple of 6, then it must also be a [#permalink]
09 Apr 2005, 19:44

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If a number is a multiple of 6, then it must also be a multiple of 3 and it must be an even number. I have a number X which is a multiple of 3. I may assume that:
A. It must be an even number.
B. It may be an even number but doesn't have to be.
C. It cannot be an even number.
D. It cannot be a multiple of 6.
E. It can be a multiple of 6 or an even number, but not both.

I dont have any OA. just wanted to confirm my answer..

A. It must be an even number - consider 9 - NO.
B. It may be an even number but doesn't have to be - pretty open-ended, but YES (consider 9, 12)
C. It cannot be an even number - consider 9 again - NO.
D. It cannot be a multiple of 6 - consider 12 - NO.
E. It can be a multiple of 6 or an even number, but not both - consider 12 again, NO.

If a number is a multiple of 6, then it must also be a multiple of 3 and it must be an even number. I have a number X which is a multiple of 3. I may assume that: A. It must be an even number. B. It may be an even number but doesn't have to be. C. It cannot be an even number. D. It cannot be a multiple of 6. E. It can be a multiple of 6 or an even number, but not both.

I dont have any OA. just wanted to confirm my answer..

A. It must be an even number.
too absolute B. It may be an even number but doesn't have to be.
moderate reasoning C. It cannot be an even number.
too absolute D. It cannot be a multiple of 6.
too absolute E. It can be a multiple of 6 or an even number, but not both.
too absolute, may be both

A. It must be an even number=> must is too strong.
B. It may be an even number but doesn't have to be=> may be ok.
C. It cannot be an even number=> cannot be too strong.
D. It cannot be a multiple of 6=> cannot be too strong.
E. It can be a multiple of 6 or an even number, but not both=> can be ok, but cannot be both means cannot be a multiple of 6 and even integer. too strong too because all multiples of 6 are even integers.

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