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# If x and y are integers, then x(x + 1)(x + 2)/(2*3*5^y) must be an int

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If x and y are integers, then x(x + 1)(x + 2)/(2*3*5^y) must be an int  [#permalink]

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10 May 2018, 01:43
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55% (hard)

Question Stats:

69% (01:30) correct 31% (01:35) wrong based on 61 sessions

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If x and y are integers, then $$\frac{x(x + 1)(x + 2)}{2*3*5^y}$$ must be an integer if which of the following is true ?

A. x is even.

B. x is odd.

C. x is divisible by three.

D. y is even.

E. y is equal to zero.

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If x and y are integers, then x(x + 1)(x + 2)/(2*3*5^y) must be an int  [#permalink]

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Updated on: 10 May 2018, 02:17
Bunuel wrote:
If x and y are integers, then $$\frac{x(x + 1)(x + 2)}{2*3*5^y}$$ must be an integer if which of the following is true ?

A. x is even.

B. x is odd.

C. x is divisible by three.

D. y is even.

E. y is equal to zero.

Questions dealing with number properties can often be solved with logic and very little calculation.
We'll look for such a solution, a Logical approach.

A fraction is an integer if its numerator is divisible by its denominator.
So, our correct answer must either say that the numerator is 0 (as 0 divided by any number is 0) or have some information about the denominator, that is about y. (Otherwise y could be any number, and specifically one that is not a factor of the numerator)
(A), (B), (C) are eliminated.
(D) is also insufficient as we can just increase y until we find a power of 5 that is not a factor of x(x+1)(x+2).

We can mark it, if we have the time we'll also see why it is true: x(x+1)(x+2) are 3 consecutive numbers and therefore at least one of them is even and one must be divisible by 3. So they are definitely divisible by 2*3 and since y = 0 means 5^y = 1, (E) is correct.
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Originally posted by DavidTutorexamPAL on 10 May 2018, 02:02.
Last edited by DavidTutorexamPAL on 10 May 2018, 02:17, edited 1 time in total.
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If x and y are integers, then x(x + 1)(x + 2)/(2*3*5^y) must be an int  [#permalink]

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10 May 2018, 02:14
Bunuel wrote:
If x and y are integers, then $$\frac{x(x + 1)(x + 2)}{2*3*5^y}$$ must be an integer if which of the following is true ?

A. x is even.

B. x is odd.

C. x is divisible by three.

D. y is even.

E. y is equal to zero.

nice question.

first we must pay attention to x, x+1, x+2..............3 consecutive integers. Any 3 consecutive integers must be divisible by 6. Look at the denominator. we 2*3 but $$5^y$$ is problematic here. Option E tell us that Y is equal to 0. $$5^0$$ = 1. thus, regardless of the value of x ultimate product will be divisible by 2*3*1.

Therefore the best answer is E.
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If x and y are integers, then x(x + 1)(x + 2)/(2*3*5^y) must be an int  [#permalink]

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10 May 2018, 04:33
Bunuel wrote:
If x and y are integers, then $$\frac{x(x + 1)(x + 2)}{2*3*5^y}$$ must be an integer if which of the following is true ?

A. x is even.

B. x is odd.

C. x is divisible by three.

D. y is even.

E. y is equal to zero.

Numerator is 3 consecutive integers. So we know the numerator is even. We also know the numerator is equal to or greater than 6. (1)(2)(3)=6. Numerator must also be divisible by 6.

Denominator is 2x3x$$5^{y}$$. If Y=0, denominator=6.

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Re: If x and y are integers, then x(x + 1)(x + 2)/(2*3*5^y) must be an int  [#permalink]

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13 May 2018, 17:55
Bunuel wrote:
If x and y are integers, then $$\frac{x(x + 1)(x + 2)}{2*3*5^y}$$ must be an integer if which of the following is true ?

A. x is even.

B. x is odd.

C. x is divisible by three.

D. y is even.

E. y is equal to zero.

We see that in the numerator, we have the product of 3 consecutive integers, which is always divisible by 3! = 6.

Thus, if y = 0, we have a denominator of 6 x 5^0 = 6 x 1 = 6, in which case the expression will always equal an integer.

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Re: If x and y are integers, then x(x + 1)(x + 2)/(2*3*5^y) must be an int &nbs [#permalink] 13 May 2018, 17:55
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