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# If positive integer y is a perfect square and is the product of r, s,

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Math Expert
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If positive integer y is a perfect square and is the product of r, s, [#permalink]

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27 Apr 2016, 11:22
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73% (00:55) correct 27% (01:32) wrong based on 174 sessions

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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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Re: If positive integer y is a perfect square and is the product of r, s, [#permalink]

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27 Apr 2016, 12:15
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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

ans to this question is B. answer explained in attachment.
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##### General Discussion
Manager
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If positive integer y is a perfect square and is the product of r, s, [#permalink]

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05 Jul 2017, 19:18
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Bunuel wrote:
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

Theory:
A perfect square has an even power of a number.
eg. 9=$$3^2$$
100=$$10^2$$

Since, y is a perfect square its factor must have an even power.
y=r.s.8.9.11
y=r.s.$$2^3$$.$$3^2$$.11
Thus, r and s must be 2.11=22
Ans: B
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Re: If positive integer y is a perfect square and is the product of r, s, [#permalink]

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24 Nov 2017, 06:41
I don't understand the solution. Because stated this way we get 2^4 * 3^2*11^2
So there are two "2" to much?

Because for a perfect square every factor should have the same exponent?
In that case I would say we have to multiply by 11^2 and 3, so rs=11^2*3 and we get 3 in every exponent.
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If positive integer y is a perfect square and is the product of r, s, [#permalink]

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25 Nov 2017, 17:19
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Bunuel wrote:
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

Dokami wrote:
I don't understand the solution. Because stated this way we get 2^4 * 3^2*11^2
So there are two "2" to much?

Because for a perfect square every factor should have the same exponent?
In that case I would say we have to multiply by 11^2 and 3, so rs=11^2*3 and we get 3 in every exponent.

Dokami This material can get confusing . . . No -- There are not enough 2s.
We need one more, to make pairs. We need four 2s, not three 2s.
There are not enough 11s either. There is only one 11. We need one more 11, so that 11s will be in pairs.

And, no: a perfect square's factors do not have the same exponent.
The prime factors must have an even exponent (e.g., 2, 6, 222).

1) Perfect square rule: A perfect square always has an even number of powers of prime factors.
That rule is often easier to remember this way:
all prime factors must come in pairs. Couplets.

There are no single "copies" of a prime factor in a perfect square.
If there is only one "copy" of a prime factor, the number is not a perfect square.

2) How the rule applies to this question
• $$y$$ "is a perfect square. and is the product of r, s, 8, 9, and 11." So

$$y = 8 * 9 * 11 * r * s$$

• Break just the numbers down into prime factors:

$$8 = (2 * 2 * 2) = 2^3$$
$$9 = (3 * 3) = 3^2$$
$$11 = (11 * 1) = 11^1$$

$$y = 2^3 * 3^2 * 11^1 * r * s$$

• Looking at just the numbers, this situation will not make a perfect square.
There are three 2s. The twos are not in pairs. (There is one pair of twos, but a third 2 is by itself.)
(2 * 2) * 2
That red 2 needs another copy of itself.

The 11 is not in a pair. There is only one 11.
If we listed out the factors as above: y = . . . $$11^1$$.

We need another 11.
We have to make pairs.
In order to make pairs, we need one more 2, and one more 11.
Both numbers need one more "copy" of themselves.

• That's what $$r$$ and $$s$$ are for.
$$r$$ and $$s$$ "make up for" the "missing" 2 and the "missing" 11.

$$(r * s) = (2^1 * 11^1) = 22$$:
(r*s) IS 22, so (r*s) must be divisible by 22.

3) Result: $$y$$ before and after
y WAS $$(2^3 * 3^2 * 11^1 * r * s)$$
We added one factor of 2 and one factor of 11 (using $$r$$ and $$s$$).

y NOW is $$2^4 * 3^2 * 11^2$$.
(2 * 2) * (2 * 2) * (3 * 3) * (11 * 11)
The factors' exponents are even numbers. We used $$r$$ and $$s$$ to make sure the factors came in pairs.
Now there are two pairs of 2s. There is one pair of 3s. And there is one pair of 11s.

Hope that helps.
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Re: If positive integer y is a perfect square and is the product of r, s, [#permalink]

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26 Nov 2017, 10:07
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Got it thank you
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Re: If positive integer y is a perfect square and is the product of r, s, [#permalink]

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23 Jan 2018, 15:12
Hi All,

This question is based on a specific Number Property rule involving perfect squares: 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.

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Re: If positive integer y is a perfect square and is the product of r, s,   [#permalink] 23 Jan 2018, 15:12
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