If positive integer y is a perfect square and is the product of r, s, 8, 9, and 11, then rs must be divisible by which of the following? (Assume both r and s are positive integers.)
A. 18
B. 22
C. 36
D. 44
E. 64
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Question on Perfect Square 7
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richachampion
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richachampion
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OA: B
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fiza gupta
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y(perfect square) = r*s*9*11*8
y = rs*9*4*2*11
9 and 4 are perfect square but not 2 and 11
to make y a perfect square rs should be 22 or multiple of 22
rs=22 or rs=22*z(z should a perfect square)(22*4,22*9......)
B is the answer
y = rs*9*4*2*11
9 and 4 are perfect square but not 2 and 11
to make y a perfect square rs should be 22 or multiple of 22
rs=22 or rs=22*z(z should a perfect square)(22*4,22*9......)
B is the answer
Fiza Gupta
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Hi richachampion,
This question is based on the same concept as the other prime-factor question that you posted: when you prime-factor a perfect square, each of the prime factors MUST show up an EVEN number of times...
eg. 9 = (3)(3) here, there are TWO 3s.
eg. 100 = (2)(2)(5)(5) here, there are TWO 2s and TWO 5s
eg. 16 = (2)(2)(2)(2) = here, there are FOUR 2s
Etc.
We're told that Y is a perfect square that is the product of R, S, 8, 9 and 11. Thus, prime-factoring Y will get us...
Y = (8)(9)(11)(R)(S)
Y = (2)(2)(2)(3)(3)(11)(R)(S)
We have TWO 3s, but only THREE 2s and just ONE 11. We need there to be an even number of 2s and an even number of 11s, so the R and S must contain at least a '2' and a '11.' Thus, (R)(S) must be a multiple of 22.
Final Answer: B
GMAT assassins aren't born, they're made,
Rich
This question is based on the same concept as the other prime-factor question that you posted: when you prime-factor a perfect square, each of the prime factors MUST show up an EVEN number of times...
eg. 9 = (3)(3) here, there are TWO 3s.
eg. 100 = (2)(2)(5)(5) here, there are TWO 2s and TWO 5s
eg. 16 = (2)(2)(2)(2) = here, there are FOUR 2s
Etc.
We're told that Y is a perfect square that is the product of R, S, 8, 9 and 11. Thus, prime-factoring Y will get us...
Y = (8)(9)(11)(R)(S)
Y = (2)(2)(2)(3)(3)(11)(R)(S)
We have TWO 3s, but only THREE 2s and just ONE 11. We need there to be an even number of 2s and an even number of 11s, so the R and S must contain at least a '2' and a '11.' Thus, (R)(S) must be a multiple of 22.
Final Answer: B
GMAT assassins aren't born, they're made,
Rich
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richachampion
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Correct Sir, I am searching and posting for other guys to practice, expertise and benefit. Thanks![email protected] wrote:Hi richachampion,
This question is based on the same concept as the other prime-factor question that you posted: when you prime-factor a perfect square, each of the prime factors MUST show up an EVEN number of times...
R I C H A,
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[email protected]
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Matt@VeritasPrep
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If we know that
r * s * 8 * 9 * 11 = (something) * (something)
we can start breaking down the numbers we have:
r * s * 2 * 2 * 2 * 3 * 3 * 11 = (something) * (something)
then trying to squeeze them into two identical square roots)
r * s * 2 * 11 = (2 * 3 * other factors) * (2 * 3 * other factors)
Since we've still got a 2 * 11 on the left side, we can put it into one of our roots:
r * s = (2 * 3 * 2 * 11) * (2 * 3 * other factors)
Since the two roots must be the same, we need
(2 * 3 * 2 * 11) = (2 * 3 * other factors)
so the other factors must contain 2 * 11. Since these other factors can only come from r and s, we must have r * s = 22 * whatever, so r * s must be divisible by 22.
r * s * 8 * 9 * 11 = (something) * (something)
we can start breaking down the numbers we have:
r * s * 2 * 2 * 2 * 3 * 3 * 11 = (something) * (something)
then trying to squeeze them into two identical square roots)
r * s * 2 * 11 = (2 * 3 * other factors) * (2 * 3 * other factors)
Since we've still got a 2 * 11 on the left side, we can put it into one of our roots:
r * s = (2 * 3 * 2 * 11) * (2 * 3 * other factors)
Since the two roots must be the same, we need
(2 * 3 * 2 * 11) = (2 * 3 * other factors)
so the other factors must contain 2 * 11. Since these other factors can only come from r and s, we must have r * s = 22 * whatever, so r * s must be divisible by 22.
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Matt@VeritasPrep
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Another approach would be to just try to make the square. Since squares have an EVEN number of each of their prime factors, we'd have to have
r * s * 2 * 2 * 2 * 3 * 3 * 11
become at least
2 * 2 * 2 * 2 * 3 * 3 * 11 * 11
To get there, we need one 2 and one 11. As before, these can only come from r * s, so r * s must contain 2 * 11 and thus be divisible by 22.
r * s * 2 * 2 * 2 * 3 * 3 * 11
become at least
2 * 2 * 2 * 2 * 3 * 3 * 11 * 11
To get there, we need one 2 and one 11. As before, these can only come from r * s, so r * s must contain 2 * 11 and thus be divisible by 22.
