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Is the positive integer n a multiple of 40?

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Is the positive integer n a multiple of 40?  [#permalink]

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New post 01 Mar 2019, 11:19
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A
B
C
D
E

Difficulty:

  75% (hard)

Question Stats:

47% (02:18) correct 53% (02:35) wrong based on 34 sessions

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Is the positive integer n a multiple of 40?

I. 20 is a factor of n^2
II. n^3/128 is an integer.
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GMAT 1: 620 Q37 V38
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Re: Is the positive integer n a multiple of 40?  [#permalink]

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New post 01 Mar 2019, 12:39
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kiran120680 wrote:
Is the positive integer n a multiple of 40?

I. 20 is a factor of n^2
II. n^3/128 is an integer.


IF YOU FIND MY SOLUTION HELPFUL, PLEASE GIVE ME KUDOS

We are asked if positive integer n is a multiple of 40

(1) 20 is a factor of n^2 20 = 2*2*5 We cannot take the square root of 5 and still have an integer, therefore min n^2 = 2*2*5*5 and min n =2*5 = 10. Therefore n = 10,20,30,40.... If n = 10 The answer is NO. if n = 40, the answer is YES NS

(2) n^3 is divisible by 128. 128 factors to 2^7. Now, in order for n to be an integer, (n^3)^1/3 must be an integer. Notice that (2^7)^(1/3) is not an integer. We need our exponent on base 2 to be divisible by 3, therefore min n^3 = 2^9 and min n=2^3 =8. If n = 8 the answer is no, but if n = 5*8 = 40 the answer is yes. NS

(1) and (2) n is a multiple of 10 and n is a multiple of 8. Therefore n is a multiple of the least common multiple of 10 and 8. 10 = 2*5 and 8 =2*2*2. Therefore n is a multiple of 2*2*2*5 = 8*5 = 40. Sufficient.

The answer is C
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Re: Is the positive integer n a multiple of 40?  [#permalink]

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New post 01 Mar 2019, 23:45
I think of these "is X a multiple of Y?" questions as: "does X contain ALL of Y's factors?"

40 = 2 x 2 x 2 x 5, or three 2's and one 5.

This is a Yes or No question.
If Yes, then "n" will have three 2's and one 5 as factors.
If No, then "n" will be account for all factors.

We don't care whether the answer is actually "yes" or "no" (no horse in this race! :)), but we just need it to be 100% certain.

1) 20 is a factor of n^2

Okay, so n^2 = n x n, and they are saying that n x n will have all the factors of 20. 20 = 2 x 2 x 5

So basically 2 x 2 x 5 would evenly divide into n x n. But hang on a minute, we have two n's in that hypothetical numerator, and the factors don't evenly split up! This tells us that "n" must have AT LEAST one 2 and one 5 as factors. So, it's possible the answer is YES, if n also has two more 2's, but what if n = 10? Then we'd get a NO answer. This is insufficient. Cross off answer choices A and D.

Let's apply the same logic to Statement 2:

2) n^3/128 is an integer.

128 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2, since we're splitting these eight 2's amongst three n's, it must be that we're missing a 2, and that each "n" has AT LEAST three 2's in it. What we don't know anything about is whether or not it has a 5. If it does, the answer is YES. If it doesn't, the answer is NO. This is insufficient; cross off answer choice B.

Combined:

"n" must have one 5 as per Statement 1, and "n" must have three 2's as per Statement 2. Therefore the answer will always be YES, that n is a multiple of 40. (Also, remember that every number is factor and a multiple of itself. So let's say n = 40. 40 is a multiple of 40, so that would give us a YES answer.)

The answer is (C).
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Re: Is the positive integer n a multiple of 40?   [#permalink] 01 Mar 2019, 23:45
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