Can any one help me with this Divisibility Query

[vatsavayi]

if 252^6 is divisible by 6^n, then what is the maximum possible value of n

 

A) 1

 

B) 2

 

C) 3

 

D) 6

 

E) 12

[Brent Hanneson]

if 252^6 is divisible by 6^n, then what is the maximum possible value of n

 

A) 1

 

B) 2

 

C) 3

 

D) 6

 

E) 12

 

A lot of integer property questions can be solved using prime factorization.

For questions involving divisibility, divisors, factors and multiples, we can say:

If N is divisible by k, then k is "hiding" within the prime factorization of N

 

Consider these examples:

24 is divisible by 3 because 24 = (2)(2)(2)(3)

Likewise, 70 is divisible by 5 because 70 = (2)(5)(7)

And 112 is divisible by 8 because 112 = (2)(2)(2)(2)(7)

And 630 is divisible by 15 because 630 = (2)(3)(3)(5)(7)

--------------------------------

Okay, onto the question:

 

We want to know how many 6's are hiding in the prime factorization of 252^6

252 = (2)(2)(3)(3)(7)

Notice that there are already two 6's hiding in the prime factorization of 252

Since 252^6 = (252)(252)(252)(252)(252)(252), we know that there must be a total of twelve 6's hiding in the prime factorization of 252^6

Answer: E

 

Cheers,

Brent

[smuhamov]

You can break 262 into prime factors as below:

262^6 = (2^2*3^2*7)^6

 

If you expand the above you get below:

2^12 * 3^12 * 7^6

 

6^n = (2 * 3) ^ n

 

If you look at prime factorization of 262 ^ 6, you have 2 ^ 12 * 3 ^ 12. As such answer should be 12.