Divisibility question

[metallicafan]

Please, check my method to analyze whether an integer is divisible by other.

Q: If x and y are positive integers, is x divisible by y?
Method:
I calculate the prime factorization of x and y
x: a^2.b.c
y: a^2.b
Being a, b, and c prime factors.
So, x/y = (a^2.b.c)/(a^2.b)
We can eliminate a^2.b in the dividend and the divisor.
So, c is the answer, and x is divisible by y.

Here, is my question:
When I cannot eliminate a prime factor of the divisor "y" (because there is not the same prime factor in the dividend "x"), in that case, x is not divisible by y, right?

Please, confirm whether my reasoning is Ok and why. Thank you very much.

[Brent@GMATPrepNow]

Please, check my method to analyze whether an integer is divisible by other.

Q: If x and y are positive integers, is x divisible by y?
Method:
I calculate the prime factorization of x and y
x: a^2.b.c
y: a^2.b
Being a, b, and c prime factors.
So, x/y = (a^2.b.c)/(a^2.b)
We can eliminate a^2.b in the dividend and the divisor.
So, c is the answer, and x is divisible by y.

Here, is my question:
When I cannot eliminate a prime factor of the divisor "y" (because there is not the same prime factor in the dividend "x"), in that case, x is not divisible by y, right?

Please, confirm whether my reasoning is Ok and why. Thank you very much.

Everything you say here is correct.

Here's how I typically handle divisibility questions:

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

For questions involving divisibility, we can say:
If N is divisible by k, then k is "hiding" within the prime factorization of N

Conversely, we can say:
If k is "hiding" within the prime factorization of N, then N is divisible by k,

Here are some examples, of what I mean when I say "hiding":
24 is divisible by 3 <--> 24 = 2x2x2x3
70 is divisible by 5 <--> 70 = 2x5x7
330 is divisible by 6 <--> 330 = 2x3x5x11
56 is divisible by 8 <--> 56 = 2x2x2x7

Cheers,
Brent