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.

Can anybody explain please?

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 aftery 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.

_________________

In the depths of winter, I finally learned

that within me there lay an invincible summer.

-- Albert Camus, "Return to Tipasa"