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multiples of 2?

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by sanju09 » Mon Jun 28, 2010 1:36 am
What is the largest number that always divides the product of three consecutive multiples of 2?
(A) 8
(B) 16
(C) 24
(D) 36
(E) 48
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by albatross86 » Mon Jun 28, 2010 2:12 am
This must ALWAYS be true. SO for any example of 3 consecutive multiples of 2, it will work.

General form is (2n)*(2n+2)*(2n+4), where n is any positive integer.

= 8*n*(n+1)*(n+2)

We know, the product of 3 consecutive integers is ALWAYS a multiple of 6. You can test cases with n(n+1)(n+2) to see this, since one of these 3 numbers must be a multiple of 3 and another a multiple of 2.

Thus the definite factor is 8*6 = 48

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by kvcpk » Mon Jun 28, 2010 2:20 am
sanju09 wrote:What is the largest number that always divides the product of three consecutive multiples of 2?
(A) 8
(B) 16
(C) 24
(D) 36
(E) 48
albatross86 is right.. IMO E
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by amising6 » Mon Jun 28, 2010 2:32 am
sanju09 wrote:What is the largest number that always divides the product of three consecutive multiples of 2?
(A) 8
(B) 16
(C) 24
(D) 36
(E) 48
three consecutive multiple of 2
first will contain minimum one 2
second will contain two 2's
third multiple will minimum contain three 2's and evey third consective even number will be multiple of 6 so one 3
so product will contain minimum six 2's i.e 2^6 and one 3
so out of given option 48 i.e 2^4*3
Ideation without execution is delusion
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