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Difficulty:
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Question Stats:
61% (02:21) correct
39%
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Is the positive integer n a multiple of 40?
(1) 20 is a factor of \(n^2\).
(2) \(\frac{n^3}{128}\) is an integer.
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(1) 20 is a factor of \(n^2\).
(2) \(\frac{n^3}{128}\) is an integer.
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Is the positive integer n a multiple of 40?
(1) 20 is a factor of \(n^2\).
(2) \(\frac{n^3}{128}\) 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
(1) 20 is a factor of \(n^2\).
(2) \(\frac{n^3}{128}\) 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
Posts: 90
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).
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).
General Discussion
Is the positive integer n a multiple of 40?
(Statement1): 20 is a factor of \(n^{2}\).
--> (n must be the product of at least \(2^{1}*5^{1}\) )
if n =10, then 20 is a factor of 100. (10 is not multiple of 40 -NO)
if n=40, then 20 is a factor of 400. (40 is a multiple of 40 -YES)
--> Insufficient
(Statement2): \(\frac{n^{3}}{128}\) is an integer.
--> In order \(\frac{n^{3}}{128}\) to be integer, n must be at least \(2^{3}\) (if it is \(2^{2}\), then (\(\frac{2^{2})^3}{128}\) -not integer )
if n= \(2^{3}\), then \(\frac{8^{3}}{128}\) -integer (8 is not multiple of 40 -NO)
if n= 40, then \(\frac{40^{3}}{128}\) - integer (40 is a multiple of 40 -YES)
Insufficient
Taken together 1&2,
n must be at least \(2^{3}*5=40\) to satisfy the both statements.
Sufficient
The answer is C.
(Statement1): 20 is a factor of \(n^{2}\).
--> (n must be the product of at least \(2^{1}*5^{1}\) )
if n =10, then 20 is a factor of 100. (10 is not multiple of 40 -NO)
if n=40, then 20 is a factor of 400. (40 is a multiple of 40 -YES)
--> Insufficient
(Statement2): \(\frac{n^{3}}{128}\) is an integer.
--> In order \(\frac{n^{3}}{128}\) to be integer, n must be at least \(2^{3}\) (if it is \(2^{2}\), then (\(\frac{2^{2})^3}{128}\) -not integer )
if n= \(2^{3}\), then \(\frac{8^{3}}{128}\) -integer (8 is not multiple of 40 -NO)
if n= 40, then \(\frac{40^{3}}{128}\) - integer (40 is a multiple of 40 -YES)
Insufficient
Taken together 1&2,
n must be at least \(2^{3}*5=40\) to satisfy the both statements.
Sufficient
The answer is C.
Posts: 73
Bunuel
Keywords: multiple, factor, divisor, divisible by
Prime factor and rephrase information as Multiple/Factor = Integer
...............
The question: n/(2^3 x 5) = Integer?
Statement (1)
n^2/(2^2 x 5) = Integer
NOTE: 5 is prime so it can’t be reduced any further!
If n^2/(2^2 x 5) = Integer —-> n^2/(5) = Integer
If n^2/(5) = Integer —-> n^2/(5^2) = Integer
Thus, Statement (1) becomes
n^2/(2^2 x 5^2) = Integer
[n/(2 x 5)]^2 = Integer
n/(2 x 5) = Integer
n = 10 —-> No (answer to original question)
n = 40 —-> Yes
Not sufficient
...........................
Statement (2) (recall “n” is a positive integer)
n^3/2^7 = Integer
(n x n x n)/(2 x 2 x 2 x 2 x 2 x 2 x 2) = Integer
In order for the above to be true, the numerator must contain at least seven 2s. But since 2 is a prime number that cannot be reduced any further, the numerator can’t possibly have exactly seven 2s. It must have at least nine 2s.
Since the numerator must have at least nine 2s:
n^3/2^9 = Integer
[n/(2^3)]^3 = Integer
n/(2^3) = Integer
Insufficient - no information about the 5 in the denominator in the question.
..........
Statement (1): n/(2 x 5) = Integer
Statement (2): n/(2^3) = Integer
Is n/(2^3 x 5) = Integer? Yes
Answer C
Posted from my mobile device
GMATRockstar
Hi,
In St.2 - you've made a mistake,
128=2^7. not 2^8
however, the answer remains unchanged.
rvgmat12
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Posts: 352
1. n^2 = 20k
n = 10 (not a multiple of 40)
n = 40 (yes)
Not sufficient
2. n^3/128 = integer
n = 8 (no)
n = 40 (yes)
1&2.
Yes. There is at least 2^3 and 5 in n
C
n = 10 (not a multiple of 40)
n = 40 (yes)
Not sufficient
2. n^3/128 = integer
n = 8 (no)
n = 40 (yes)
1&2.
Yes. There is at least 2^3 and 5 in n
C
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