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Question on Perfect Square 7

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by richachampion » Mon Oct 17, 2016 3:11 am
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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by richachampion » Mon Oct 17, 2016 3:11 am
OA: B
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by fiza gupta » Mon Oct 17, 2016 4:57 am
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
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by [email protected] » Mon Oct 17, 2016 10:00 am
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

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by richachampion » Mon Oct 17, 2016 10:37 am
[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...
Correct Sir, I am searching and posting for other guys to practice, expertise and benefit. Thanks!
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by Matt@VeritasPrep » Fri Oct 28, 2016 12:38 am
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.
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by Matt@VeritasPrep » Fri Oct 28, 2016 12:39 am
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.
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