Does this rule about factors and exponents always hold true?

[Guest]

I took this from an explanation from the Official Guide (don't worry I won't post the actual problem).

The explanation states that "because x^3 is divisible by 9, x must be divisible by 3".

Why is this true?

My assumption was that, since 3^2 is a factor of x^3 (and x is the only source of factors), then 3 must be a factor of x. But does this ALWAYS hold true? Can you say, since x^9 is divisible by 9, than x must be divisible by 3?

Thanks,
mdh

[RonPurewal]

I took this from an explanation from the Official Guide (don't worry I won't post the actual problem).

The explanation states that "because x^3 is divisible by 9, x must be divisible by 3".

Why is this true?

My assumption was that, since 3^2 is a factor of x^3 (and x is the only source of factors), then 3 must be a factor of x. But does this ALWAYS hold true? Can you say, since x^9 is divisible by 9, than x must be divisible by 3?

Thanks,
mdh

yes, on all counts. in fact, it doesn't even have to be 9; even if all we know is that x^n (for any n > 1) is divisible by 3, that's enough to deduce that x is divisible by 3.
the reason is because x^n is the product of n identical "x"s, each of which has the same prime factors. therefore, for any given prime factor, there are only two choices:
* that prime factor appears in the prime factorization of x (and thus appears in x^n at least n times)
* that prime factor doesn't appear in the prime factorization of x (and thus doesn't appear in x^n at all)
if there are any 3's in the prime factorization of x at all, then that means 3 goes into x, which means that the prime factorization of x^n actually contains at least n "3"s.

note that this only HAS to be true for PRIME numbers, though. for composite numbers, it may or may not be true.
for instance, if you know that x^3 is divisible by 64, it does NOT follow that x is divisible by 8. however, if x^3 is divisible by 16, then x DOES have to be divisible by 4. (see if you can work out the details; if not, post back)

[mdh3000]

note that this only HAS to be true for PRIME numbers, though. for composite numbers, it may or may not be true.
for instance, if you know that x^3 is divisible by 64, it does NOT follow that x is divisible by 8. however, if x^3 is divisible by 16, then x DOES have to be divisible by 4. (see if you can work out the details; if not, post back)

Thanks for the help Ron!

Here is my thinking....

If x^3 is divisible by 64, then we know x^3 has the prime factors 2^6. However, the best we can say is that each x has 1/3 of those prime factors, so 2^2 = 4. So we can say x is divisible by 4, but not by 8.

If x^3 is divisible by 16, then we know x^3 has the prime factors 2^4. Evenly distributed, two x's have 2 as a factor and one x has 2^2 as a factor. However, since each x is equal, we know that each x must have 2^2 = 4 as a factor. And thus x is divisible by 4.

Is my thinking correct?

mdh

[rfernandez]

Absolutely correct. Good work and a good question.

[Guest]

yes, on all counts. in fact, it doesn't even have to be 9; even if all we know is that x^n (for any n > 1) is divisible by 3, that's enough to deduce that x is divisible by 3.


What if X = the square root of 3 and N = 2?

[RonPurewal]

note that this only HAS to be true for PRIME numbers, though. for composite numbers, it may or may not be true.
for instance, if you know that x^3 is divisible by 64, it does NOT follow that x is divisible by 8. however, if x^3 is divisible by 16, then x DOES have to be divisible by 4. (see if you can work out the details; if not, post back)

Thanks for the help Ron!

Here is my thinking....

If x^3 is divisible by 64, then we know x^3 has the prime factors 2^6. However, the best we can say is that each x has 1/3 of those prime factors, so 2^2 = 4. So we can say x is divisible by 4, but not by 8.

If x^3 is divisible by 16, then we know x^3 has the prime factors 2^4. Evenly distributed, two x's have 2 as a factor and one x has 2^2 as a factor. However, since each x is equal, we know that each x must have 2^2 = 4 as a factor. And thus x is divisible by 4.

Is my thinking correct?

mdh

your thinking is correct.
in the latter part, note that you're right: each of the x's must have two 2's in its prime factorization. this means that if the perfect cube is divisible by 16, then, in fact, it must be divisible by 2^6 = 64.

well played!

[RonPurewal]

yes, on all counts. in fact, it doesn't even have to be 9; even if all we know is that x^n (for any n > 1) is divisible by 3, that's enough to deduce that x is divisible by 3.


What if X = the square root of 3 and N = 2?

i believe the original discussion pertained to integers only. in any case, official problems will not ask you to address concerns of divisibility, remainders, and so on, unless the numbers in question are positive integers. so, while you are technically correct here, you should learn to frame these arguments around numbers that are assumed to be integers.

note that this is not general advice; it is a bad idea to assume that ALL numbers on the gmat are integers. indeed, you should check the problem statement of any problem - including divisibility problems - carefully, just to ensure that the variables really must be integers; on many problems (especially data sufficiency problems), there will be no such stipulation.

[guest]

Just to continue...I want to verify:

If x^3 is divisible by 32 then x is divisible by 2, 4, and 8?

32=2^5

The 2s can be evenly distributed among the x's in the following way:

2 2^2 and 2^2
or
2 2 and 2^3


Thanks.

[guest]

Just to continue...I want to verify:

If x^3 is divisible by 32 then x is divisible by 2, 4, and 8?

32=2^5

The 2s can be evenly distributed among the x's in the following way:

2 2^2 and 2^2
or
2 2 and 2^3


Thanks.

Example:
If x^3 is divisible by 32 then x must be divisible by...2

So does this example mean that x must be divisible by 2 and not necessarily by 4 or 8 based on the logic in the quote above? Thanks in advance for your help.

[RonPurewal]

Just to continue...I want to verify:

If x^3 is divisible by 32 then x is divisible by 2, 4, and 8?

32=2^5

The 2s can be evenly distributed among the x's in the following way:

2 2^2 and 2^2
or
2 2 and 2^3


Thanks.

um, well, evenly distributed means evenly distributed.
so:
the minimum requirement is the first one you've mentioned above (2, 2^2, 2^2), you see, if any of the x's contains 2^2, then all of them have to (because each x is the same number!).
it's obviously impossible to distribute five 2's without having 2 of them in one place, so this is the absolute minimum. therefore, x must be divisible by 4 (and therefore also by 2), but NOT necessarily by 8.

(if you have 2, 2^2, 2^3, then you have to put three 2's in each x.)

by example, the smallest perfect cube divisible by 32 is 64, which is divisible by 4 (= 2^2) but not by 8 (= 2^3). that example alone disproves the notion that x must be divisible by 2^3.

[guest]

Thanks for the clarification.

[rfernandez]

You're welcome!