IMO : D

Question Stem:\((a^2)(b^2)\) = 400

\((a^2)(b^2) = (2^4)(5^2)\)

a*b = \((2^2)(5^1)\)

a,b = (20,1) or (4,5) or (2,10)
Thus 3 pairs.

First break down 200 into 20*20 and further into the prime factors 2*2*5*2*2*5. Now we are looking for all the possible pairs (2 numbers) of squares whose product results in 400.

1st: 2^2*10^2 (i.e. the first two 2's and two times 2*5 = 10)
2nd: 4^2*5^2 (i.e. two times 2*2 = 4 = 4^2 and 5^2).
3rd: 1^2*20^2 (i.e. two times 2*2*5 and 1^2 = 1)

Answer D.

Given : (x^2)*(y^2) = 400

i.e. (x*y)^2 = (20)^2

i.e. (x*y) = (20)

1*20 = 20
2*10 = 20
4*5 = 20

i.e. 3 Possible pairs [OR Total Possible pairs = No. of Factors of (20=2^2*5)/2 = (2+1)(1+1)/2 = 6/2 = 3]

Answer: option D
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Ans: D - 3 pairs

(xˆ2)(yˆ2) = 400 [square root both sides]

xy = 20

20 = 1x20, 2x10, 4x5, 5x4, 10x2, 20x1

Cancel the repeats

This leaves us with exactly 3 options.

Hence, D

\(x^2 * y^2 = 400\)

\((xy)^2 = 400\)

\(xy = 20\)

\(x\) and \(y\) can take the following values: \((x,y) = (1,20), (2,10), (4,5)\). Hence, Ans (D).
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800score Official Solution:

400 = 2 × 2 × 2 × 2 × 5 × 5
Combine the prime factors in pairs.

400 = (2 × 2) × (2 × 2) × (5 × 5)
Now brake the factorization into two parts, each one will be a square.
The possible combinations are:
400 = (2 × 2) × [(2 × 2) × (5 × 5)]
400 = [(2 × 2) × (2 × 2)] × (5 × 5)
But don't forget that 400 = 1 × 400, where 1 = 1². So we also have:
400 = (1 × 1) × [(2 × 2) × (2 × 2) × (5 × 5)]

Thus all the possible combinations of the factors that make the product of two squares are the following:
1² × 20² = 400
2² × 10² = 400
4² × 5² = 400

There are three possible pairs that fit the criterion. The correct answer is D.
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Can we reduce the expression to x*y=20 or x*y=5*2^2
And then approach it like this? The number of factors in 20 are 6, since we are looking for pair of factors, we divide de total number of factor by 2 and get 3 as a result.

Is this a right reasoning?

We can express 400 as a product of two numbers:

400 = 1 x 400 = 2 x 200 = 4 x 100 = 5 x 80 = 8 x 50 = 10 x 40 = 16 x 25 = 20 x 20

We see that of the products above, three products are products of two perfect squares:

1 x 400 = 1^2 x 20^2

4 x 100 = 2^2 x 10^2

16 x 25 = 4^2 x 5^2

Thus, there are 3 pairs of positive integers that satisfy the condition.

Alternate Solution:

If we denote the integers as n and m, we are given that (n^2)(m^2) = 400, which can be rewritten as (nm)^2 = 400 and simplified as nm = 20.

Thus, for any two integers whose product is 20, we get a pair of integers that satisfy the given condition. There are three such pairs: 20 x 1 = 10 x 2 = 5 x 4 = 20.

Answer: D
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lets say 2 numbers are x and y.

So the equation is , x^2 * y^2 = 400
xy = 20

20 can be written as 1X20, 2X10, and 4X5, so 3 ways.

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