If x and y are prime numbers, how many factors has x^2y^2?

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[Math Revolution GMAT math practice question]

If x and y are prime numbers, how many factors has x^2y^2?

1) xy=10
2) x+y is odd

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by fskilnik@GMATH » Tue Aug 21, 2018 11:05 am

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<p>
</p>
If x and y are prime numbers, how many factors has x^2y^2?

1) xy=10
2) x+y is odd
We may assume without loss of generality that x is not greater than y (due to the symmetries in the question stem and in our focus).

More than that, each statement alone is sufficient to guarantee that x and y are different. (Why?)

Conclusion: 2 <= x < y , both primes (in each statement alone).

This is enough to guarantee the answer is unique (and the same) in each statement alone (*), hence the answer is D.

(*) It is (2+1)*(2+1) if only positive factors are considered, twice this value in reality (therefore 18).

(Some people consider "factor" as a "positive divisor". This is wrong in math in general, and also in GMAT in particular, as a careful look at the Official Guide shows.)

Kudos to the proposer of this problem: He/She did not violate an important "Data Sufficiency rule": you cannot find a unique answer with the question stem pre-statements (only).
In fact, before the given statements, the primes x and y COULD be equal and, in this case, the answer would be 10 (= 2*(4+1)). In other words, pre-statements, we have 10 and 18 as "potential answers".
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by GMATGuruNY » Tue Aug 21, 2018 11:16 am

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To count the positive factors of a positive integer:
1. Prime-factorize the integer
2. Write the prime-factorization in the form (a^p)(b^q)(c^r)...
3. The number of factors = (p+1)(q+1)(r+1)...

Generally, GMAT problems about factors are constrained to POSITIVE factors.
On the GMAT, the posted problem would probably appear as follows:
Max@Math Revolution wrote:[Math Revolution GMAT math practice question]

If x and y are prime numbers, how many POSITIVE factors has x^2y^2?

1) xy=10
2) x+y is odd
Only two cases are possible:
Case 1: x≠y
Here, to determine the number of factors for x²y², we add 1 to each exponent and multiply:
(2+1)(2+1) = 9 factors
Case 2: x=y, with the result that x²y² = x²x ²= x�
Here, to determine the number of factors for x�, we add 1 to the exponent:
4+1 = 5 factors

Implication:
To determine the number of factors, we need to know whether x=y.
Question stem, rephrased:
Does x=y?

Statement 1:
Only one pair of prime numbers has a product of 10:
2 and 5.
Thus, x≠y.
SUFFICIENT.

Statement 2:
Since x+y = odd, either x or y is ODD, while the other value is EVEN.
Thus, x≠y.
SUFFICIENT.

The correct answer is D.
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by Max@Math Revolution » Thu Aug 23, 2018 1:04 am

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=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question.

If x and y are different prime numbers, then xy has (2+1)(2+1) = 9 factors.
If x and y are the same prime number, then xy has 4+1 = 5 factors.

Condition 1)
Since x and y are prime numbers and xy = 10, either x = 2 and y = 5, or x = 5 and y = 2.
So, x and y are different prime numbers. Thus, condition 1) is sufficient.

Condition 2)
Since x and y are prime numbers and x + y is odd, one of them is even and the other one is odd.
So, x and y are different prime numbers. Thus, condition 2) is sufficient.

Therefore, D is the answer.

Answer: D