The positive integer q is divisible by 15. If the product of

Question

The positive integer q is divisible by 15. If the product of q and the positive integer x is divisible by 20, which of the following must be a factor of  x^2*q ?

A. 25
B. 60
C. 75
D. 150
E. 300

KarishmaB · Accepted Answer

q = 3*5*n (n is some integer)
x*q = 4*5*m (m is some integer)
Since q has 3 as a factor, x*q must have 3 as a factor too.
x*q = 3*4*5*a = 60*a (a is some integer)
Hence 60 must be a factor of x*q.

What about  x^2*q ? It must have x*x*q = x*60*a
So, 60 must be a factor of  x^2*q .

Notice that all the other options have 25 as a factor. We do not know whether 5 is a factor of x or not. Hence the opther options COULD be factors of  x^2*q , but it is not essential that they have to be. Only 60 MUST be a factor of  x^2*q .

The cheat code of every GMAT question is the same: Have a strong conceptual understanding!

GyanOne · Answer

q is divisible by 15 and the product of q and x is divisible by 20
=> x is a multiple of 4

Therefore x^2(q) will have two fours and a 15 in its product, i.e. it will have 60, 16, 15, and 240 as factors.

Of the given options, only B fits into these possibilities. B is therefore the answer.

skiingforthewknds · Answer

Is this right, I feel like I got it by scrambling and plugging in numbers.

q/15 and qx/20

for the equations to fit I plugged in 30 for q (satisfying the first equation) and 2 for x in the second 60 (2*30)/20

so x=2 and q=30

(2^2)*30 = 120......60 is the only one that goes into it.

Is this right with my logic?

Thank you.

Bunuel · Answer

q  could  be 30 and x  could  be 2. In this case options A, C, D and E are not the factors of x^2q, thus these options are NOT ALWAYS true. Thus, by POE (process of elimination) the answer must be B.

Hope it's clear.

msharmita · Answer

Hello, Please explain this in a simple manner. did not get the solution

mittalg · Answer

Given: q is divisible by 15, and x*q is divisible by 20

q = 15 k (minimum value of q = 15)

When x*q is divide 20, we get a 5 from q. We need a 4 to make it divisible by 20. Therefore, minimum values of x =4

or x= 4n

x^2 * q = 16 n^2 * 15 K = 240 r

This is divisible by 60 among the options given.

Also note that we have only one 5 coming from 15 which we can be sure of. All other options are factors of 25 which can't divide some number containing single 5.

Hope you get it!!!

Please press kudos if you like!!!

KarishmaB · Answer

I think you need to go through the basics of factors. Check out my post: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2014/02 ... r-factors/ Then come back to the solution. It might make more sense.

Lucky2783 · Answer

Q=3*5*k
xq divisible by 20 so x must have atleast 4

X2q will have atleast 4×4×3×5 . Clearly 60 must be a factor of x2q

EMPOWERgmatRichC · Answer

Hi Pretz,

This question can be solved by TESTing VALUES. Since the prompt asks which of the following MUST be a factor, we have to TEST the smallest values possible.

This prompt gives us a number of facts to work with:
1) Q is a positive integer divisible by 15
2) X is a positive integer
3) (Q)(X) is divisible by 20

IF....
Q = 15
X = 4
(Q)(X) = 60

We're asked which of the following must be a factor of (X^2)(Q)?

(X^2)(Q) = (4^2)(15) = (16)(15) = 240

The only answer that is a factor is 60.

Final Answer:  B

GMAT assassins aren't born, they're made,
Rich

suramya26 · Answer

q=15k
q=15,30,45,60...
q*x=20p
Hence 
q*x=20,40,60....
Let's say x=1(for convenience)
Then common value is 60
q=60
x^2*q= 60x^2
Hence 60 will always be a factor....

Focus on concept not on question

Posted from my mobile device

Abhishek009 · Answer

METHOD - I 
 
PLUG IN SOME VALUES FOR q

CASE - I

q = 15 { Completely divisible by 15 }

x = 4

qx = 60 { Completely divisible by 20}

CASE - II

q = 30 { Completely divisible by 15 }

x = 6

qx = 180 { Completely divisible by 20}

So, In each case the number is divisible by 60....

METHOD - II

q/15 = Rem 0

So, qx/15*20 = Rem 0

Check for the least possible number that will divide qx without any remainder using 15 & 20

qx = { 60, 120 , 180..........}

Hence answer will be (B) 60

ra5867 · Answer

What level question this must be?
Thanks

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Bunuel · Answer

Check the stats in the original post for tags and other stats: 2018-01-06_0008.png

NDND · Answer

I solved this problem with the LCM (Least Common Multiple of two numbers is the smallest number that is a multiple of both)

15= 5*3
 20= 5*3*2*2
LCM= 5*3*2*2 = 60 (B)

orangebicycle5 · Answer

I break this up into three sections:

1. Make q = the primes of 15, so 3*5. qx = the primes of 20, so 2^2 * 5. The 5's will cancel out but we need a 4 or 2^2 to make 20 divisible by qx.

2. x = (2^2)^2 * 3 * 5 --> 2^4 * 3 * 5

3. Look at answer choices, put into primes and see which cancel the primes in step 2. The primes of 60 = 2^2*3*5, so this would make x an integer.

DanTheGMATMan · Answer

Parse x away from qx and then set up the rest:

https://www.youtube.com/watch?v=DgBiAeD9HGA

bumpbot · Answer

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