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Is the product of four consecutive integers a multiple of

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Is the product of four consecutive integers a multiple of [#permalink] New post 09 Jul 2006, 03:13
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A
B
C
D
E

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Is the product of four consecutive integers a multiple of 120?

(1) None of the four digits is a multiple of 10.
(2) The sum of the four integers is a multiple of 10.
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 [#permalink] New post 10 Jul 2006, 07:39
What kind of explanation is that?
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 [#permalink] New post 10 Jul 2006, 08:04
B.
interesting question

(1) is not sufficient. Note: 120=2*2*2*3*5, for example 23*24*25*26 is a multiple of 120, or 120*121*122*123.
(2) is sufficient.
We know that the last digits of the four numbers must be either 1, 2, 3 and 4 (1+2+3+4=10) or 6, 7, 8, 9 , these numbers constructed in that way
(i. e. k*10+1, k*10+2, K*10+3, k*10+4, k integer)
are not divisible by 5 (analogously for 6, 7, 8, 9). Hence, We know that these numbers cannot be multiples of 120.

Therefore, B is the correct answer.

Last edited by game over on 10 Jul 2006, 08:41, edited 1 time in total.
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 [#permalink] New post 10 Jul 2006, 08:06
I have a feeling it should be (C), but can prove this completely

Let's write this as n(n+1)(n+2)(n+3) divisible by 120

Stem1. Let's pick 1,2,3,4 as for consecutive integers. Their product is 24, not divisible by 120.

On the other hand, 2*3*4*5 is divisible by 120.

/NOT SUFF/

Stem2. Pick the same 1,2,3,4 because their sum is equal to 10. Again, NOT SUFF

I think when you combine two statements, it should be enough to prove that the product is NOT divisible by 120, but I don't want to spend my time on it right now
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 [#permalink] New post 10 Jul 2006, 08:10
After reading game over's solution, I agree it should be (B).
Thanks!
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 [#permalink] New post 10 Jul 2006, 09:02
Great insight!!

Took me a while to get it, but finally got it.

I was going with E as I couldn't get anywhere with the Statement 2.

Thanks/

game over wrote:
B.
interesting question

(1) is not sufficient. Note: 120=2*2*2*3*5, for example 23*24*25*26 is a multiple of 120, or 120*121*122*123.
(2) is sufficient.
We know that the last digits of the four numbers must be either 1, 2, 3 and 4 (1+2+3+4=10) or 6, 7, 8, 9 , these numbers constructed in that way
(i. e. k*10+1, k*10+2, K*10+3, k*10+4, k integer)
are not divisible by 5 (analogously for 6, 7, 8, 9). Hence, We know that these numbers cannot be multiples of 120.

Therefore, B is the correct answer.
  [#permalink] 10 Jul 2006, 09:02
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