If x and y are integers and xy = 660, x or y must be divisible by whic

Question

If x and y are integers and xy = 660, x or y must be divisible by which of the following?

A. 3 
B. 4 
C. 6 
D. 20 
E. 30

Kurtosis · Answer

xy = 660 = 2 * 2 * 3 * 5 * 11

660 is even and a multiple of 3 --> So x or y must be a multiple of 3.

Let's check it out:
Consider 660 = 2 * 330 --> x or y cannot be a multiple of 4, 20
Consider 660 = 4 * 165 --> x or y cannot be a multiple of 6, 30

Answer: A

chetan2u · Answer

Vyshak  a good solution..
The reasoning is that the solution has to be a PRIME number because any other composite number can be expressed in terms which may not necessarily contain that composite number..
example 36 may be 6*6 BUT 36 can still be expressed in PRODUCT of factors that may not contain..
36 = 4*9..

so a solution would bea prime factor of 660..
here 3 is given 
A

mathiaskeul · Answer

Just using prime factorization or normal factorization will easily tell you its going to be a multiple of 3

330 times 2 55 times 12 etc

dcummins · Answer

I don't completely understand why it needs to be divisible by 3 and why it can't be divisible by 4 i.e. 2^2, which is also contained in the prime factorisation.

Anyone able to clarify?

jibberjabber · Answer

I cannot seem to understand why 3 is the answer. Can someone please explain it to me?

Subhrajyoti · Answer

The prime factorization of 660 = 3*2*2*5*11, so we can use permutaions here for making x and y, i see that 4, 6,3 and 20 can divide, so why 3 is the right answer. Hoping to get a better explaination.

chetan2u · Answer

For the members above who have the query on this question.

xy =660, so x and y can be any factors of 660.

Now the question is dealing with MUST, so the integer should always be factor of x or y in all possible combinations.

Ok, here we can write 660 as product of its factors and find if there is ANY combination that does not contain the integer. 
A. 3......660=2*330=11*60=10*60.... In all the cases there will be one of x or y that will be a multiple of 3.
B. 4.....Now 660 has 2^2 in it so if I keep one 2 in x and other in y, there will be a case, where neither x nor y is multiple of 4. 660=66*10=2*330
C. 6.... If I keep all 2s in x and 3s in y, we will have neither x nor y divisible by 6. 660=4*165=20*33=220*3
D. 20.. If you have understood the reasoning for 4 and 6, this is even more straightforward. 660=66*10
E. 30..,Again 660=66*10

A. 
Of course 2 is also an answer but it is not given in choices. 
Say the number were 660*2=1320, Now 4 will also be the answer as 1320 contains 2^3, and we can divide this at the best in 2 and 2^2, so one would surely be divisible by 2^2 or 4.

akara2500 · Answer

How about if x is 1 and y is 660, then x isn't divisible by 3?

chetan2u · Answer

The question says x or y is divisible by ??
So, it becomes ‘1 or 660’ is divisible by 3? Yes, 660 is.

Priyankjain5 · Answer

beacuse 4 is not a prime number like 33 multiply by 20 is 660 it is not a divisible by 4 but its by 3

If x and y are integers and xy = 660, x or y must be divisible by whic

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If x and y are integers and xy = 660, x or y must be divisible by whic

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