x,y are positive integers. Find the number of even factors of 4*x^2.
I. x^3-y^3+3xy is odd and x is a prime.
II. x^(x+y) * y^(3x) + x^(3y) is odd.
OA : C
@ Experts - could you please share your explanation to solve this sort of problems in most optimized way within (2 mins)time constraint ? Much thanks in advance!
x,y are positive integers. Find the number of even factors
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Since an ODD integer raised to a positive integer power stays ODD, and an EVEN integer raised to a positive integer power stays EVEN, the exponents in the two statements are irrelevant and can be disregarded.RBBmba@2014 wrote:x,y are positive integers. Find the number of even factors of 4*x^2.
I. x^3-y^3+3xy is odd and x is a prime.
II. x^(x+y) * y^(3x) + x^(3y) is odd.
Statement 1: x - y + 3xy is odd and x is a prime.
Case 1: y=1
Then x - y + 3xy is odd becomes:
x - 1 + 3x = odd
4x = odd + 1
4x = even.
Case 1a: x=2
Here, 4x² = 16, which has even factors 2, 4, 8 and 16.
Case 1b: x=3
Here, 4x² = 36, which has even factors 2, 4, 6, 12, 18 and 36.
Since the number of even factors can be different values, INSUFFICIENT.
Statement 2: xy + x is odd
Case 1: y=1
Here, xy + x = odd becomes:
2x = odd.
Not possible.
Case 2: y=2
Here, xy + x = odd becomes:
3x = odd.
Here, it is possible that x=3 (Case 1b).
In Case 1b, 4x² = 36, which has even factors 2, 4, 6, 12, 18 and 36.
Case 2a: x=1
In this case, 4x² = 4, which has even factors 2 and 4.
Since the number of even factors can be different values, INSUFFICIENT.
Statements combined:
Only Case 1b satisfies both statements, implying that x must be an ODD PRIME NUMBER.
Thus, 4x² = 2²(odd prime)², with the result that 4x² will always yield the SAME NUMBER OF EVEN FACTORS.
SUFFICIENT.
The correct answer is C.
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Much thanks GMATGuru. I did it it in the same way but got confused at the end and it resulted in high solving time...What should be the EXPECTED time within which we're supposed to solve this type of problems ? (I took around 3.5 mins or so)
To determine this I chose x to be 3 and 5 and both the cases found that NUMBER OF EVEN FACTORS to be same(SIX). However, this trial method consumed a lot time than anticipated, hence would like to know how we can establish logically that if x is an ODD PRIME NUMBER then 4x² will always yield the SAME NUMBER OF EVEN FACTORS ?
Look forward to know your thoughts!
I chose 9 and ended up consuming more time here...missed out that 1 is a non-prime odd. (here, I believe I could save some good amount of time) Any thoughts how one should keep an open-eye to such smart choices of plug-in numbers ?GMATGuruNY wrote: Statement 2: xy + x is odd
.
.
.
Case 2a: x=1
In this case, 4x² = 4, which has even factors 2 and 4.
GMATGuruNY wrote:
Statements combined:
Only Case 1b satisfies both statements, implying that x must be an ODD PRIME NUMBER.
Thus, 4x² = 2²(odd prime)², with the result that 4x² will always yield the SAME NUMBER OF EVEN FACTORS.
To determine this I chose x to be 3 and 5 and both the cases found that NUMBER OF EVEN FACTORS to be same(SIX). However, this trial method consumed a lot time than anticipated, hence would like to know how we can establish logically that if x is an ODD PRIME NUMBER then 4x² will always yield the SAME NUMBER OF EVEN FACTORS ?
Look forward to know your thoughts!
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Mitch - could you please share your thoughts on my IMMEDIATE above post ?
Much thanks in advance!
Much thanks in advance!
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I suspect that many test-takers would take 2-3 minutes to solve this problem.RBBmba@2014 wrote:Much thanks GMATGuru. I did it it in the same way but got confused at the end and it resulted in high solving time...What should be the EXPECTED time within which we're supposed to solve this type of problems ? (I took around 3.5 mins or so)
Always test the SMALLEST POSSIBLE CASE.I chose 9 and ended up consuming more time here...missed out that 1 is a non-prime odd. (here, I believe I could save some good amount of time) Any thoughts how one should keep an open-eye to such smart choices of plug-in numbers ?GMATGuruNY wrote: Statement 2: xy + x is odd
.
.
.
Case 2a: x=1
In this case, 4x² = 4, which has even factors 2 and 4.
I evaluated statement 2 by testing y=1 and x=1, the smallest possible options for x and y.
Number of even factors = (total number of factors) - (total number of odd factors).GMATGuruNY wrote:b]Statements combined:[/b]
Only Case 1b satisfies both statements, implying that x must be an ODD PRIME NUMBER.
Thus, 4x² = 2²(odd prime)², with the result that 4x² will always yield the SAME NUMBER OF EVEN FACTORS.
To determine this I chose x to be 3 and 5 and both the cases found that NUMBER OF EVEN FACTORS to be same(SIX). However, this trial method consumed a lot time than anticipated, hence would like to know how we can establish logically that if x is an ODD PRIME NUMBER then 4x² will always yield the SAME NUMBER OF EVEN FACTORS ?
Look forward to know your thoughts!
I explain how to count the total number of factors and the total number of odd factors here:
https://www.beatthegmat.com/gmat-loves-f ... 72876.html
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If you have a prime factorization, written in the normal way (using exponents), the power on the 2 always tells you the ratio of the number of even factors to the number of odd factors. So for example, if you have this number:
2^5 * 3^4 * 7^3
then because '5' is the power on the 2, the ratio of even to odd factors will be 5 to 1. In other words, 1/6 of this number's factors are odd, and 5/6 of its factors are even. You can easily see why this is true: if you take any odd factor of this number, you can make exactly five even factors by multiplying by 2, 2^2, 2^3, 2^4 and 2^5.
You probably know that to count a number's factors, we add 1 to each exponent in the prime factorization, and multiply what we get. Here the exponents are 5, 4, and 3. If we add one to each of these, we get 6, 5, and 4. So the number above has 6*5*4 = 120 factors. Since 1/6 of these are odd, 20 of the factors are odd, and 100 are even.
So in the question you posted above, if we reach the conclusion that our prime factorization looks like:
2^2 * p^2
where p is an odd prime, then we know that the ratio of even to odd factors is 2 to 1, so 2/3 of this number's factors are even. Using the method to count factors, this number has (2+1)(2+1) = 9 factors in total, and (2/3)(9) = 6 of them are even.
2^5 * 3^4 * 7^3
then because '5' is the power on the 2, the ratio of even to odd factors will be 5 to 1. In other words, 1/6 of this number's factors are odd, and 5/6 of its factors are even. You can easily see why this is true: if you take any odd factor of this number, you can make exactly five even factors by multiplying by 2, 2^2, 2^3, 2^4 and 2^5.
You probably know that to count a number's factors, we add 1 to each exponent in the prime factorization, and multiply what we get. Here the exponents are 5, 4, and 3. If we add one to each of these, we get 6, 5, and 4. So the number above has 6*5*4 = 120 factors. Since 1/6 of these are odd, 20 of the factors are odd, and 100 are even.
So in the question you posted above, if we reach the conclusion that our prime factorization looks like:
2^2 * p^2
where p is an odd prime, then we know that the ratio of even to odd factors is 2 to 1, so 2/3 of this number's factors are even. Using the method to count factors, this number has (2+1)(2+1) = 9 factors in total, and (2/3)(9) = 6 of them are even.
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Thanks Brent! Hope you're doing well.
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