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Difficulty:
95%
(hard)
Question Stats:
27% (02:58) correct
73%
(02:33)
wrong
based on 44
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If x is an integer, does 2x have twice as many factors as x does?
(1) \(x^y\) is divisible by 2 where y is an integer
(2) \(\frac{x}{y}\) is even
(1) \(x^y\) is divisible by 2 where y is an integer
(2) \(\frac{x}{y}\) is even
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Archit3110
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Updated on: 18 Oct 2021, 07:22
Originally posted by Archit3110 on 16 Oct 2021, 09:24.
Last edited by Archit3110 on 18 Oct 2021, 07:22, edited 2 times in total.
Last edited by Archit3110 on 18 Oct 2021, 07:22, edited 2 times in total.
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given x is an integer target does 2*x have twice as many factors as x
#1
\(x^y\) is divisible by 2 where y is an integer
irrespective of the value of y value of x would always be even
so x=2,4,6,8,10,12,16
factors of x for
2=2
4=3
6=4
8=4
10=4
12=6
16= 5
2x for these values ; 4,8,12,16,24,32
factors of 2x would be
4=3
8=4
12=6
24=8
32=6
for all the cases 2x does not have twice as many factors as x does
or x=0 factors of 0 are infinite so 2x can be true as factors of twice as many factors of 0
insufficient
#2
\(\frac{x}{y}\) is even
irrespective of the value of y ; x has to be an even integer
again same as #1 2x does not have twice as may factors as x
or x=3 and y=1/2 ; in this case we get yes as factors of 3 =2 and factors of 2*3 =6 is 4
we get yes and no so #2 is insufficient
from 1 &2 we cannot determine anything conclusively
option E is correct
#1
\(x^y\) is divisible by 2 where y is an integer
irrespective of the value of y value of x would always be even
so x=2,4,6,8,10,12,16
factors of x for
2=2
4=3
6=4
8=4
10=4
12=6
16= 5
2x for these values ; 4,8,12,16,24,32
factors of 2x would be
4=3
8=4
12=6
24=8
32=6
for all the cases 2x does not have twice as many factors as x does
or x=0 factors of 0 are infinite so 2x can be true as factors of twice as many factors of 0
insufficient
#2
\(\frac{x}{y}\) is even
irrespective of the value of y ; x has to be an even integer
again same as #1 2x does not have twice as may factors as x
or x=3 and y=1/2 ; in this case we get yes as factors of 3 =2 and factors of 2*3 =6 is 4
we get yes and no so #2 is insufficient
from 1 &2 we cannot determine anything conclusively
option E is correct
GMATinsight
UPDATED 10.17
Statement 1 -- Sufficient.
X to the power of y is divisible by 2. For any number to be a multiple of 2, it must be even. And if Y is an integer, X to the power of Y would need to be an even perfect square. X could be positive or negative since any number squared is positive. So test cases for X could be positive or negative.
Some possible values for x and 2x are...
x=-4, -2, 0, 2, 4, 6, 8
2x=-8, -4, 0, 4, 8, 12, 16
If X is -4, X has 5 factors--1, -1, -2, 2, 4, -4--and 2x has 8 factors -- 1, -1, 2, -2, 4, -4, 8, -8. Answer is no, not double.
If X is 0, X has one factor--0--and 2x equals 0, which also has 1 factor. Answer is no, not double.
If X is 4, X has 3 factors--1, 2, 4--and 2X has 4 factors--1, 2, 4, 8. Again, answer is no, not double.
No matter what permissible value of X, 2X never has double the number of factors that X has. Statement 1 is sufficient.
Statement 2 -- Insufficient.
"x/y is even" tells us a few things about x. X could be even, and that would yield the same 'no' as above. But, for this statement, we don't know anything about Y to suggest it must be an integer, so X could also be odd, which can violate this statement. For example, if x/y is 6, X could be 3 and Y could be 1/2. 3 (X) has two factors--1, 3--whereas 6 (2x) has 4 factors--1, 2, 3, 6. This yields a "yes". X could be 4 and Y could be 2, making the answer to the question stem "no", because 2x has 3 factors whereas X has 2 factors. So, statement 2 is insufficient.
The correct answer is A.
Statement 1 -- Sufficient.
X to the power of y is divisible by 2. For any number to be a multiple of 2, it must be even. And if Y is an integer, X to the power of Y would need to be an even perfect square. X could be positive or negative since any number squared is positive. So test cases for X could be positive or negative.
Some possible values for x and 2x are...
x=-4, -2, 0, 2, 4, 6, 8
2x=-8, -4, 0, 4, 8, 12, 16
If X is -4, X has 5 factors--1, -1, -2, 2, 4, -4--and 2x has 8 factors -- 1, -1, 2, -2, 4, -4, 8, -8. Answer is no, not double.
If X is 0, X has one factor--0--and 2x equals 0, which also has 1 factor. Answer is no, not double.
If X is 4, X has 3 factors--1, 2, 4--and 2X has 4 factors--1, 2, 4, 8. Again, answer is no, not double.
No matter what permissible value of X, 2X never has double the number of factors that X has. Statement 1 is sufficient.
Statement 2 -- Insufficient.
"x/y is even" tells us a few things about x. X could be even, and that would yield the same 'no' as above. But, for this statement, we don't know anything about Y to suggest it must be an integer, so X could also be odd, which can violate this statement. For example, if x/y is 6, X could be 3 and Y could be 1/2. 3 (X) has two factors--1, 3--whereas 6 (2x) has 4 factors--1, 2, 3, 6. This yields a "yes". X could be 4 and Y could be 2, making the answer to the question stem "no", because 2x has 3 factors whereas X has 2 factors. So, statement 2 is insufficient.
The correct answer is A.
