[demoskopico]

p^a q^b r^c s^d= x, where x is a perfect square. If p, q, r, and s are prime integers, are they distinct?

(1) 18 is a factor of ab and cd
(2) 4 is not a factor of ab and cd

please very slow and careful explanations. thanks

[JohnB]

In this question, p, q, r, and s represent the prime factors of x, and the exponents a, b, c, and d tell us how many times each of those factors appears. Since x is a perfect square, each of its prime factors appears an even number of times.

(Why is this? The square root of x is equal to the product of the square roots of its prime factors--the square root of the product is always the same as the product of the square roots. In order for the square root of x to be an integer, then, the product of the square roots of its prime factors must be an integer. But no prime number is a perfect square. So each square root of a prime factor must have a match in order to yield an integer at the end.

For example, the perfect square 144 = 2*2*2*2*3*3. Sqrt (144) = sqrt(2) *sqrt(2) *sqrt(2) * sqrt(2) * sqrt (3) * sqrt (3) = 2*2*3. We end up with an integer because all the square roots pair up.

On the other hand, 160 is not a perfect square. 160 = 2*2*2*2*2*5. Sqrt (160) = sqrt(2)*sqrt(2)*sqrt (2)*sqrt (2) *sqrt(2) * sqrt (5) = 4sqrt(10). The result is not an integer because we have odd prime factors floating around.)

Now that we see we need an even number of prime factors for x, the question whether p, q, r, and s are distinct reduces to the question, are a, b, c, and d all even? If so, then p, q, r, and s could be distinct (although they don't have to be).

If one of a, b, c, or d is odd, then the only way to get an even number of occurrences of the corresponding prime factor is to have it reappear. For instance, if a = 3 and p = 2, then at least one of q, r, and s must also = 2 and have an odd exponent as well. 2^3*3^2*2^1*5^2 is a possible value for x (it would equal 3600).

Now let's look at our choices.

I. ab and cd each have a factor of 18. This doesn't tell us much. We could have a = b = c = d = 6, in which case p, q, r, and s maybe are distinct and maybe are not. INSUF.

II. ab and cd do NOT have a factor of 4. Then one of a or b and one of c or d must be odd. Otherwise, each would have a factor of 2, and their products would have a factor of 4. From our reasoning above, if one of the exponents is odd, then p, q, r, and s CANNOT be all distinct. SUF.

So answer is B.

I realize that this is a fairly involved explanation--please let me know if you have questions--it would be best if you could point out specifically any steps that are unclear.

[demoskopico]

Originally Posted by JohnB

In this question, p, q, r, and s represent the prime factors of x, and the exponents a, b, c, and d tell us how many times each of those factors appears. Since x is a perfect square, each of its prime factors appears an even number of times.

(Why is this? The square root of x is equal to the product of the square roots of its prime factors--the square root of the product is always the same as the product of the square roots. In order for the square root of x to be an integer, then, the product of the square roots of its prime factors must be an integer. But no prime number is a perfect square. So each square root of a prime factor must have a match in order to yield an integer at the end.

For example, the perfect square 144 = 2*2*2*2*3*3. Sqrt (144) = sqrt(2) *sqrt(2) *sqrt(2) * sqrt(2) * sqrt (3) * sqrt (3) = 2*2*3. We end up with an integer because all the square roots pair up.

On the other hand, 160 is not a perfect square. 160 = 2*2*2*2*2*5. Sqrt (160) = sqrt(2)*sqrt(2)*sqrt (2)*sqrt (2) *sqrt(2) * sqrt (5) = 4sqrt(10). The result is not an integer because we have odd prime factors floating around.)

Now that we see we need an even number of prime factors for x, the question whether p, q, r, and s are distinct reduces to the question, are a, b, c, and d all even? If so, then p, q, r, and s could be distinct (although they don't have to be).

If one of a, b, c, or d is odd, then the only way to get an even number of occurrences of the corresponding prime factor is to have it reappear. For instance, if a = 3 and p = 2, then at least one of q, r, and s must also = 2 and have an odd exponent as well. 2^3*3^2*2^1*5^2 is a possible value for x (it would equal 3600).

Now let's look at our choices.

I. ab and cd each have a factor of 18. This doesn't tell us much. We could have a = b = c = d = 6, in which case p, q, r, and s maybe are distinct and maybe are not. INSUF.

II. ab and cd do NOT have a factor of 4. Then one of a or b and one of c or d must be odd. Otherwise, each would have a factor of 2, and their products would have a factor of 4. From our reasoning above, if one of the exponents is odd, then p, q, r, and s CANNOT be all distinct. SUF.

So answer is B.

I realize that this is a fairly involved explanation--please let me know if you have questions--it would be best if you could point out specifically any steps that are unclear.

ok that's very clear. anyway, I believe I wouldn't be able to come up with this in just 2 minutes duing the test. is it normal or am too bad in math???