If q is a integer, is q^4 a multiple of 64?
(1) q^4 is not a multiple of 128.
(2) q^2 has 27 factors, 7 of which are less than or equal to 10
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is q^4 a multiple of 64?
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question - whether q^4 is a multiple of 64 or not
a.) This is insufficient. Prove by quoting 2 instances. 192 is not a multiple of 128 yet, it is a multiple of 64. Also, 130 is also not a multiple of 128 and is not a multiple of 64 either. Hence this information is definitely not sufficient.
b.) This is also insufficient. 7 of the 27 factors are equal to or less than 10. If they had been exactly 10, then, you would have (5x2) repeating 7 times over, thus making up for 2^7 = 128 (a multiple of 64). In this case q^4 would have (128x128) as a factor and hence would also be a multiple of 64. But if the 7 factors are say 9 (3x3) and say the other 20 factors are also not a multiple of 2, then q^4 would not be a multiple of 64. Hence the information is not suffiecient.
c.) The two statements together also do not provide any relevant or coherent information. Hence the answer is also not [C]
I think, hence, the answer should be [E]
a.) This is insufficient. Prove by quoting 2 instances. 192 is not a multiple of 128 yet, it is a multiple of 64. Also, 130 is also not a multiple of 128 and is not a multiple of 64 either. Hence this information is definitely not sufficient.
b.) This is also insufficient. 7 of the 27 factors are equal to or less than 10. If they had been exactly 10, then, you would have (5x2) repeating 7 times over, thus making up for 2^7 = 128 (a multiple of 64). In this case q^4 would have (128x128) as a factor and hence would also be a multiple of 64. But if the 7 factors are say 9 (3x3) and say the other 20 factors are also not a multiple of 2, then q^4 would not be a multiple of 64. Hence the information is not suffiecient.
c.) The two statements together also do not provide any relevant or coherent information. Hence the answer is also not [C]
I think, hence, the answer should be [E]
If q is Odd...then q^4 will be odd. So (q^4/64) and (q^4/128) can never be integers. No specific conclusion from this.
If q is Even.
(q^4/128) = (q^4/64) * (1/2).
The only way (q^4/128) can be an integer is IF (q^4/64) is an even integer.
I am unable to find a case where (q^4/64) is an odd integer.
So from the above two statements, if (q^4/128) is not an integer, then (q^4/64) is also NOT an integer.
If q is Even.
(q^4/128) = (q^4/64) * (1/2).
The only way (q^4/128) can be an integer is IF (q^4/64) is an even integer.
I am unable to find a case where (q^4/64) is an odd integer.
So from the above two statements, if (q^4/128) is not an integer, then (q^4/64) is also NOT an integer.
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netigen
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Since, q is an integer it will have distinct prime factors. Lets assume for a moment that 2 is a factor of q (if 2 is not a factor of q then 64 can never be a factor of q^4)
in that case for q^4 the power of two will be a multiple of 4 i.e. the factor will of the form 2^(4k)
from A we know that 128 is not a factor of q^4 hence we know that 2^7 is not a factor of q^4
which means 4k < 7 or k = 1
so, maximum possible power of 2 factor for q^4 can be 2^4 = 16 hence we know that 64 will never be a factor of q^4 if 128 is not it's factor therefore A is sufficient
for (B) we know that there are 7 factors below 11
if we pick up all possible even factors 1,2,3,4,6,8,10 then q^4 becomes a factor of 64
but if we pick these factors 1,2,3,5,6,7,9 then q^4 will not be a factor of 64 so both cases possible hence insufficient.
This is the best approach I could figure out for this question.
in that case for q^4 the power of two will be a multiple of 4 i.e. the factor will of the form 2^(4k)
from A we know that 128 is not a factor of q^4 hence we know that 2^7 is not a factor of q^4
which means 4k < 7 or k = 1
so, maximum possible power of 2 factor for q^4 can be 2^4 = 16 hence we know that 64 will never be a factor of q^4 if 128 is not it's factor therefore A is sufficient
for (B) we know that there are 7 factors below 11
if we pick up all possible even factors 1,2,3,4,6,8,10 then q^4 becomes a factor of 64
but if we pick these factors 1,2,3,5,6,7,9 then q^4 will not be a factor of 64 so both cases possible hence insufficient.
This is the best approach I could figure out for this question.
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amitansu
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This q baffled me enough !! I concluded 'E' as the ans.
But again i found it was my mistake.Had this been a true GMAT DS prob i would have definitely succumbed !!
From stem 2 : it'sevident that it is not sufficient.
From stem 1: My understanding says :
some no. ,which is in the form of to the power 4 of its , is not divisible by 128.
Now 128 is a multiple of 64.So any no. which is in the form of "t the 4th power of its" and is not divisible by 128 , is definitely not divisible by 64.
So q^4 is not a multiple of 64.
Ans A.
But again i found it was my mistake.Had this been a true GMAT DS prob i would have definitely succumbed !!
From stem 2 : it'sevident that it is not sufficient.
From stem 1: My understanding says :
some no. ,which is in the form of to the power 4 of its , is not divisible by 128.
Now 128 is a multiple of 64.So any no. which is in the form of "t the 4th power of its" and is not divisible by 128 , is definitely not divisible by 64.
So q^4 is not a multiple of 64.
Ans A.
where is this question from?
1) q^4 is not a multiple of 128
to be a multiple of 64, q^4 needs to be a multiple of 2^6.
if q^4 is not a multiple of 128, then q^4 does not contain 7 2's in its factorization. Since we can only go down in sets of 4 of any factor (since we have q^4), if q^4 contains any 2's, it must contain either 2^4, 2^8, etc, since we know it contains less than 2^7, we can conclude it doesn't contain 2^6 either (since 2^4 is the next possible number of 2s)
1 is sufficient
2) q^2 has 27 factors, 7 of which are less than or equal to 10
to show this is insuff, we need to show 2 cases, one in which 2) is met and is a multiple of 64, and one in which 2) is met and is not a multiple of 64
q^2 has 27 factors, with 7 less than 10, means that we could have the 7 factors all be 2's, in which case we know it is a factor of 64 (regardless of the other 20 factors).
On the otherhand, if all 7 factors are 3's (and the other 20 factors are 11's), then we know it is not a multiple of 64.
2 is insuff
Therefore the answer is A
1) q^4 is not a multiple of 128
to be a multiple of 64, q^4 needs to be a multiple of 2^6.
if q^4 is not a multiple of 128, then q^4 does not contain 7 2's in its factorization. Since we can only go down in sets of 4 of any factor (since we have q^4), if q^4 contains any 2's, it must contain either 2^4, 2^8, etc, since we know it contains less than 2^7, we can conclude it doesn't contain 2^6 either (since 2^4 is the next possible number of 2s)
1 is sufficient
2) q^2 has 27 factors, 7 of which are less than or equal to 10
to show this is insuff, we need to show 2 cases, one in which 2) is met and is a multiple of 64, and one in which 2) is met and is not a multiple of 64
q^2 has 27 factors, with 7 less than 10, means that we could have the 7 factors all be 2's, in which case we know it is a factor of 64 (regardless of the other 20 factors).
On the otherhand, if all 7 factors are 3's (and the other 20 factors are 11's), then we know it is not a multiple of 64.
2 is insuff
Therefore the answer is A
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s_raizada
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Just to simplyfy
Let's used substituion to solve this
q (integer) q4 Multiple of 64
2 8 No
3 21 No
4 256 Yes but it is also a multiple of 128 so violates the constraint in 1
5 625 NO
6 1296 no
7 xxx No
8 4096 yes but it is also multiple of 128 so violates the constraint in 1
Therefore q4 is not a multiple of 64
Let's used substituion to solve this
q (integer) q4 Multiple of 64
2 8 No
3 21 No
4 256 Yes but it is also a multiple of 128 so violates the constraint in 1
5 625 NO
6 1296 no
7 xxx No
8 4096 yes but it is also multiple of 128 so violates the constraint in 1
Therefore q4 is not a multiple of 64
