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# If q is a integer, is q^4 a multiple of 64? (1) q^4 is not a

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If q is a integer, is q^4 a multiple of 64? (1) q^4 is not a [#permalink]

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17 Sep 2006, 04:21
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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.
VP
Joined: 21 Aug 2006
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17 Sep 2006, 04:44

2^8*3^2 is then number..

only in that case 7 factors (1,2,3,4,6,8,9) are less than or equal to 10.
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Re: DS: Multiple of 64 [#permalink]

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17 Sep 2006, 04:49
kevincan wrote:
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.

Stem: q^4 must contain a minimum of six 2s (2^6=64)

1) q^4 doesn't contain 7 2s (2^7) - insufficient.

2) prime numbers <10 are 2,3,5,7. the 7 factors can be six 2s and one 3 (fits), or six 3s and one 2 (then q^4 has only 2 twos - doesn't fit), or no 2s at all. insufficient

combined: there are no more than seven 2s in q^4, but we din't know the lower limit.

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17 Sep 2006, 05:10
I get d)

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

From 1)

There are no integer values of q so that q^4 is not both a multiple of 128 or 64. Therefor SUFF

From 2)

If try and make it so that q^4 is not a multiple of 64, ie picking numbers so as not to creat 2^6 (ie 64) in the factorisation we get the factors below 10:
2*3*5*6*7*9*10 ---> 6 = 2*3 and 10 = 2*5

Squaring this and looking at the factors of 2 to give n^4 gives 2^2 * 2^2 * 2^2 = 2^6 or 64

So q^4 is a multiple of 64

Only exception would be if you can include 1 as a factor below 10 - is this allowed?
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17 Sep 2006, 05:13
I'm also assuming the factors less than or equal to 10 have to be different
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17 Sep 2006, 05:32
I think A) has to be sufficient because:

Question asks if N^4/64 is an integer where N is an integer.

64 = 2^4 * 2^2

For N to be an integer the 4th root of the factors must also be integers so minimum value of factors is 2^4 * 2^4 or 128 * 2 or 64 * 2^2

Therefore A must be SUFF
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Re: DS: Multiple of 64 [#permalink]

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17 Sep 2006, 06:29
1
KUDOS
kevincan wrote:
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.

This is really a good one folks, I am sure the answer is D (either alone is sufficient)

Folks, we will learn a new concept here, so it's time to concentrate again.

Is q^4 a multiple of 64?
Ok if q^4 is a multiple of 64 it should be equal to 64xk
ie q^4 = 64 x k (64 x k must be a raised to power 4)
=>q^4 = 2^4 x 2^2 x k.

So minimum value of k must be 2^2.
So minimum value of q^4 = 2^4 x 2^2 x 2^2
Hence the minimum value of q^4 = 2^8 = 256, which is divisble by 128.

So we can conclude that any number raised to power 4, if it is divisible by 64 then it will be definitely divisible by 128.
But statement 1, says that q^4 is not divisible by 128

So we can conclude that q^4 is not divisible by 64.
Hence Statement 1 alone is sufficient.

Let's look at statement 2:
q^2 has 27 factors, 7 of which are less than or equal to 10.

I think all of know the way to calculate the total number of factors.
Anyhow I am explaining it here.

We split the number into prime factors say
N = P1^a1 x P2^a2 x .................. x Pn^an
Total number of factors = (a1+1) x (a2+1) x........x(an+1).

Now it is given that q^2 has 27 factors. Let's look at different cases here.

Case 1: q^2 = P^26 (in this case total factors are 27)
So min value of q^2 = 2^26, but we have only 1,2,4,8 factors which are less than 10. So this is not valid.
Since the min value do not contain 7 factors less than 10, obviously number greater than this will not have 7 factors less than 10

Case 2: q^2 = P1^2 x P2^8 ( in this case total factors = 3 x 9 = 27)
So min value of q^2 = 2 ^2 x 3^8
Now the factors of this number less than 10 are 1,2,3,4,6,9 only. So even this is out........

Case 3:q^2 = P1^2 x P2^2 x P3^2 (in this case totalfactors = 3x3x3=27)
So min value of q^2 = 2^2 x 3^2 x 5^2
Now the factors of this number less than or equal to 10 are 1,2,3,5,4,6,8,9,10 which are infact more than 7.
So even this is out.
So going for the next value of q^2
q^2 = 2^2 x 3^2 x 7^2
Now the factors of this number less than or equal to 10 are
1,2,3,7,4,6,9 which are exactly 7.
So min value of q = 2x3x7.

So finally this is the number that satisfies statement 2, and
clearly q^4 is not divisible by 64.

After there is no need to verify other numbers because any other number will not have 7 facotrs less than or equal to 10 and contains 27 factors.

Hence statement 2 alone is sufficient

So folks, finally the answer is D.

a bit tiring isn't it? But believe me u have to learn concepts like these, so that 700+ is sureshot.

One more thing here is if we understand the above concept believe me u can solve any question based on these factors in our GMAT.

Finally, this post seems to be big, don't be afraid please, to make everything lucid, I have explained every case possible ..................

Huh, I need some rest now................
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Last edited by cicerone on 25 Sep 2008, 01:23, edited 1 time in total.
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17 Sep 2006, 09:43
Cicerone, why do you talk about min. value when evaluating (2)? Were you too quick to dismiss Case 2?
VP
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18 Sep 2006, 06:23
kevincan wrote:
Any more thoughts?

How abt this?

Case 2: q^2 = P1^2 x P2^8 ( in this case total factors = 3 x 9 = 27)
So min value of q^2 = 3 ^2 x 2^8
Now the factors of this number less than 10 are 1,2,3,4,6,8,9 only.

The number is divisible by 64. No?

Whether it is divisible or not, then answer seems to be D.
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18 Sep 2006, 06:39
I dont know...but to me this is E...

q^4=64*M..? is the question...in other words does q^4= have 7 2s.

1) q^4 is not a multiple of 128...

well q^4=64, which is not a multiple of 128 but is definetly a multiple of 64...q^4=65 which is not a multiple of 128 and is not a mutliple of 64..Insuff

2) q^2= has 7 factors.. they can 1,2,3,4,5,6,7, which has 4 2's in it..then q^4 will have 8 2's ...in this case q^4 will be a multiple of 64...

or q^2=1,3,5,6,7,9,10, in this case there are only 2 2's..q^4 will then only have 4 2's...which is not a multiple of q^4....Insuff

E it is...
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19 Sep 2006, 07:59
ak_idc wrote:
kevincan wrote:
Any more thoughts?

How abt this?

Case 2: q^2 = P1^2 x P2^8 ( in this case total factors = 3 x 9 = 27)
So min value of q^2 = 3 ^2 x 2^8
Now the factors of this number less than 10 are 1,2,3,4,6,8,9 only.

The number is divisible by 64. No?

Whether it is divisible or not, then answer seems to be D.

I don't understand this, ak_idc
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19 Sep 2006, 10:19
fresinha12 wrote:
I dont know...but to me this is E...

q^4=64*M..? is the question...in other words does q^4= have 7 2s.

1) q^4 is not a multiple of 128...

well q^4=64, which is not a multiple of 128 but is definetly a multiple of 64...q^4=65 which is not a multiple of 128 and is not a mutliple of 64..Insuff

E it is...

64 has 6 2s

and from 1) If q is an integer then q^4 must contain factors raised to the power of 4. 64 = 2^4 * 2^2

and hence 128 = 2^4 * 2^2 * 2 or 2^7 which you cannot take a fourth root of yielding an integer value of q

Therefore if 2 is a factor then minimum that q^4 could equal is 2^4 = 16 (not a multiple of 64)

or 2^4 * 2^4 = 2^4 * 2^4 = 2^8 = 128 * 2 or 64 * 4

Therefore if q is not a multiple of 128 it cant be a multiple of 64 and a is SUFF
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19 Sep 2006, 20:21
kevincan wrote:
ak_idc wrote:
kevincan wrote:
Any more thoughts?

How abt this?

Case 2: q^2 = P1^2 x P2^8 ( in this case total factors = 3 x 9 = 27)
So min value of q^2 = 3 ^2 x 2^8
Now the factors of this number less than 10 are 1,2,3,4,6,8,9 only.

The number is divisible by 64. No?

Whether it is divisible or not, then answer seems to be D.

I don't understand this, ak_idc

Oh..sorry..what I meant was Case 2 can satisfy the conditions. I mean it can have two prime factors (2,3), can have 27 total factors, and also have 7 factors of this number less than 10.

And the minimum value of the number is 2304 or 3^2*2^8, and this number is divisible by 64.

Cicerone concluded that the number is not divisible by 64.
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20 Sep 2006, 05:26
For those of you non-math inclined, here's another way to look at problems with this type of wording. Some of you eluded to it in your posts, but I'll spell it out...

Original Post:
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.

Reworded:
If q is an integer, is (q^4)/64 an integer?

1) (q^4)/128 is not an integer
2) Thanks for simplifying this Cicerone

From 1) you can easily see that it's sufficient as 128 = 64 x 2 , and you can avoid the E answer trap

A few more:
"y is a multiple of x" is saying, "y/x is an integer"

or I've seen this in some study books, 'Is 2x+2 a multiple of y?" This is asking, "Is (2x + 2)/y an integer?"
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20 Sep 2006, 10:28
I have been watching this topic for long time and still don't understand it.

Kevin, please correct me where I'm wrong.

Quote:
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.

How 1 stmt can be sufficient?

It says only that q^4 is not multiple of 128. Fine. q^4 can be 64*3 or 64*5. From that q^4 is multiple of 64. But q^4 can be 127, and in this can q^4 is multiple of 64. INSUF !!!

Stmt2.

q^2 has 27 factors. Lets find prime factors.

Lets q^2= x^n*y^k*z^p, where x,y,z are primes and n,m,p are integers. Total number of factors are (n+1)(m+1)(p+1)=27=3*3*3

From that q^2 has three prime factors.

7 of that factors less or equal to 10. From that we find can find that q^2 has 2,3,5 as it's prime factors. 1,2,4,5,6,9,10 are first 10 factors.

q^2=2^2*3^2*5^2

From that q^4=2^4*3^4*5^4

In order q^4 be multiple of 64 it has to have 2^6 as its factor.

So stmt 2 is SUFF to tell that q^4 is NOT multiple of 64.

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20 Sep 2006, 11:14
Natalya,

You're missing a condition from the stem in statement 1)

The stem says n IS AN INTEGER and asks if n^4 is divisible by 64.

So for n^4 to satisfy the stem it has to yield an integer when you take the fourth root ie fourth root of 16 = 2.

Now 128 = 2^7 (so you cant take a fourth root and get integer value factors - 7 is not divisible by 4)

So if if n = 1 n^4 = 1 - not divisible by 64

If n = 2 then 2^8 = 256 = 128 * 2 = 64 * 2^2

For the fourth root of n^4 to equal an integer than the factors have to be raised to a power that is divisible by 4.

Now 64 = 2^6 which cant be a value of n^4 and so would have to be n^8 which, as shown above, is both a multiple of 128 and 64. If you carry on with n^12, n^16 ....... they will all be multiples of 128 AND 64 and so statement 1 is sufficient to say that n^64 is not a multiple of 64.
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Re: DS: Multiple of 64 [#permalink]

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20 Sep 2006, 14:48
kevincan wrote:
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.

(1) As q is an integer, when we write q as a product of distinct prime factors, the power of 2 must be a whole number (non-negative integer). Thus the power of 2 in the prime factorization of q^4 must be a multiple of 4. As 2^7 is not a factor of q^4, the highest power of 2 that could be a factor of q^4 is 2^4. Thus q^4 is not a multiple of 64=2^6. SUFF

(2) q^2 has 27 factors. Thus q^2 could be of the form (i) a^26, (ii)b^2*c^8 or (iii) a^2*b^2*c^2 where a b and c are distinct prime numbers.
But as q^2 has 7 factors less than 11, it is obvious that (i) is impossible.

As for (ii) 2^8*3^2 has exactly 7 factors under 11 (1,2,3,4,6,8,9), in which case q^4 would be a multiple of 64. 3^8*2*2 is not a possibility for q^2 (it only has 6 factors under 1: 1,2,3,4,6,9)

Regarding (iii) if q^2=2^2*3^2*5^2, q^2 would have 8 factors under 11- 1,2,3,4,5,6,9,10 and q^2=2^2*3^2*7^2 would have 7 factors under 11: 1,2,3,4,6,7,9) In this case, q^4 would not be a multiple of 64. Thus (2) is insufficient.

OA=A. I hope all of you liked this question. Will you find a question like this in the exam? Only to separate a 51 from a 50, possibly. Keep in mind that Q-talented people from Asia are doing the exam in increasing numbers, and people have more resources to prepare for the GMAT than I did in my day (how bad that sounds, my day) I envy all of you because you are all young enough to do an MBA- how exciting it will be!

That said, the thinking exercise involved in this question is helpful for all of us.
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Re: DS: Multiple of 64 [#permalink]

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21 Sep 2006, 07:07
i totally forgot q was an integer...

kevincan wrote:
kevincan wrote:
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.

(1) As q is an integer, when we write q as a product of distinct prime factors, the power of 2 must be a whole number (non-negative integer). Thus the power of 2 in the prime factorization of q^4 must be a multiple of 4. As 2^7 is not a factor of q^4, the highest power of 2 that could be a factor of q^4 is 2^4. Thus q^4 is not a multiple of 64=2^6. SUFF

(2) q^2 has 27 factors. Thus q^2 could be of the form (i) a^26, (ii)b^2*c^8 or (iii) a^2*b^2*c^2 where a b and c are distinct prime numbers.
But as q^2 has 7 factors less than 11, it is obvious that (i) is impossible.

As for (ii) 2^8*3^2 has exactly 7 factors under 11 (1,2,3,4,6,8,9), in which case q^4 would be a multiple of 64. 3^8*2*2 is not a possibility for q^2 (it only has 6 factors under 1: 1,2,3,4,6,9)

Regarding (iii) if q^2=2^2*3^2*5^2, q^2 would have 8 factors under 11- 1,2,3,4,5,6,9,10 and q^2=2^2*3^2*7^2 would have 7 factors under 11: 1,2,3,4,6,7,9) In this case, q^4 would not be a multiple of 64. Thus (2) is insufficient.

OA=A. I hope all of you liked this question. Will you find a question like this in the exam? Only to separate a 51 from a 50, possibly. Keep in mind that Q-talented people from Asia are doing the exam in increasing numbers, and people have more resources to prepare for the GMAT than I did in my day (how bad that sounds, my day) I envy all of you because you are all young enough to do an MBA- how exciting it will be!

That said, the thinking exercise involved in this question is helpful for all of us.
Re: DS: Multiple of 64   [#permalink] 21 Sep 2006, 07:07
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