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If b is the product of three consecutive positive integers

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Eternal Intern
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If b is the product of three consecutive positive integers [#permalink] New post 17 Jul 2003, 11:29
24. If b is the product of three consecutive positive integers c, c + 1, and c + 2, is b a multiple of 24 ?

(1) b is a multiple of 3,
(2) c is odd.

The answer is A, it cannot be right
B=60 when C= 3 Answer is No
B=336 when C= 6 Answer is Yes
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 [#permalink] New post 17 Jul 2003, 14:28
The answer should be B. (1) doesn't tell you anything since all multiples of 3 consecutive ints are a multiple of 3.
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AkamaiBrah
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MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993

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Re: VT [#permalink] New post 17 Jul 2003, 14:39
Curly05 wrote:
It's E


You're right. Now can you tell us why?
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AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
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MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993

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 [#permalink] New post 17 Jul 2003, 19:41
In 2:

Subsitute the values 1 & 15. One is divicible and the other is not.

Thus, Answer is E.

Is there a better mathematical representation to it?
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 [#permalink] New post 17 Jul 2003, 22:21
kpadma wrote:
In 2:

Subsitute the values 1 & 15. One is divicible and the other is not.

Thus, Answer is E.

Is there a better mathematical representation to it?


Yes. We need 2*2*2*3 to make 24. There will always be a factor of 3 in a sequence of 3 numbers. Since both the first and last numbers are odd, we cannot get any contributions of 2's from them. Hence, only when the middle number is a factor of 8 will the product be divisible by 24, but that is not necessarily the case so (2) it is not sufficient.

My mistake was that I concentrated on the EVEN side. Any sequence starting with an even number is ALWAYS divisible by 24. Why? because every sequence of 3 has one number divisible by 3. In addition, both end numbers will be divisible by 2 with one of them always divisible by 4, hence you have all of the necessary factors for any even starting number. Knowing this, I simple assumed that all of the odds were NOT true, but I neglected to recognize the case where the middle number is divisible by 8.

Is that clear enough?
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AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993

  [#permalink] 17 Jul 2003, 22:21
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