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The answer should be B. (1) doesn't tell you anything since all multiples of 3 consecutive ints are a multiple of 3.
_________________

Best,

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

You're right. Now can you tell us why?
_________________

Best,

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

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?
_________________

Best,

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