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If x and y are positive integers and 1620x/y^2 is the square of an odd 19 Apr 2019, 12:23
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C
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45% (medium)

Question Stats:

71% (02:27) correct 29% (02:33) wrong based on 798 sessions

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If x and y are positive integers and \(\frac{1620x}{y^2}\) is the square of an odd integer, what is the smallest possible value of xy?

A) 1
B) 8
C) 10
D) 15
E) 28

My solution:
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1620x/y² = (Odd#)²

√(1620x/y²) = Odd#

here I factor 1620 = 162*10 -> 81*2²*5 -> 9²*2²*5*x

9*2*√(5*x) /y = Odd#

E/E and O/O can both give Odd# ints, but since there's a 2 in the numerator, it means E/E ... so y has to be even

The smallest even int is 2, so y = 2

Now to remove the sqroot in the numerator x must be at least 5, so the minimum value is 2*5 = 10
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C
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If x and y are positive integers and 1620x/y^2 is the square of an odd 19 Apr 2019, 16:28
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\(\frac{1620x}{y^2} = z^2\), with z being an odd integer(or \(2k+1\), if you fancy)

Having in mind that \(\sqrt{\frac{1620x}{y^2}}\) is an integer, we can easily see that \(√y^2 = y\) and now we need to factor \(1620x\) to find an x that makes \(\frac{1}{y}\sqrt{1620x}\) also integer:

\(1620x = 3^4*2^2*5*x\)

Now we have \(\frac{3^2*2√5x}{y}\) integer. Having in mind that √5x must also be an integer, we are looking for the smallest possible integer x (since we are also looking for the smallest xy) that makes the square root integer, which is x=5.

With that we find \(\frac{90}{y} = 2k+1 (odd)\),

So, to turn an even number, such as \(90\), into an odd one we need to divide it by another even number, which makes:

\(y=2k (even)\) , but...

Hey! We are also looking for the the smallest product for xy, so y must be the smallest even number, which is 2.

Finally,
\(x = 5\) and \(y = 2\), so \(xy = 10\)

Answer: C
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If x and y are positive integers and 1620x/y^2 is the square of an odd 25 Jan 2020, 11:29
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Quick solution - Reduce 1620 into factors - 2*5*2*81
Now writing it in square form - 2^2 * 9^2 * 5 x/y^2
x has to be 5 for numerator to form a perfect square. For xy to be minimum, choose value of y which is the least such that 1620x/y^2 is an integer which gives y = 2. Hence xy = 10
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If x and y are positive integers and 1620x/y^2 is the square of an odd 17 Jan 2020, 08:07
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energetics
If x and y are positive integers and \(\frac{1620x}{y^2}\) is the square of an odd integer, what is the smallest possible value of xy?

A) 1
B) 8
C) 10
D) 15
E) 28
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E/E=even, odd, fraction, undefined
O/O=odd, fraction
E/O=even, fraction
O/E=undefined, fraction

\(\frac{1620x}{y^2}=odd^2…odd^2=odd*odd\)

\(1620x=even…\frac{even}{y^2}=odd^2…y^2=even…(\frac{E}{E}=odd)\)

\(1620=162*10=81*2*2*5=3^42^25\)

\(\frac{1620x}{y^2}=odd^2=perf.square…powers(1620x)=even\)

\(1620x=3^42^25x…minimum(x)=5…min(y=even)=2…min(xy)=10\)

Ans (C)
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If x and y are positive integers and 1620x/y^2 is the square of an odd 24 Jul 2020, 05:34
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joaopschultz
\(\frac{1620x}{y^2} = z^2\), with z being an odd integer(or \(2k+1\), if you fancy)

Having in mind that \(\sqrt{\frac{1620x}{y^2}}\) is an integer, we can easily see that \(√y^2 = y\) and now we need to factor \(1620x\) to find an x that makes \(\frac{1}{y}\sqrt{1620x}\) also integer:

\(1620x = 3^4*2^2*5*x\)

Now we have \(\frac{3^2*2√5x}{y}\) integer. Having in mind that √5x must also be an integer, we are looking for the smallest possible integer x (since we are also looking for the smallest xy) that makes the square root integer, which is x=5.

With that we find \(\frac{90}{y} = 2k+1 (odd)\),

So, to turn an even number, such as \(90\), into an odd one we need to divide it by another even number, which makes:

\(y=2k (even)\) , but...

Hey! We are also looking for the the smallest product for xy, so y must be the smallest even number, which is 2.

Finally,
\(x = 5\) and \(y = 2\), so \(xy = 10\)

Answer: C
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If x and y are positive integers and 1620x/y^2 is the square of an odd 29 Jul 2020, 23:27
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This is how i approached this question.

the square of an odd integer will be an odd integer. To become an odd integer there can be no even number multiplied in the numerator.

now if we Reduce 1620 into factors - 2*5*2*81 = 2^2 * 9^2 *5
so to become an odd perfect square , we have to cancel the 2^2 and multiply by 5 .
thus if the minimum value of y = 2 and x = 5, all these can be achieved.
so the minimum value of xy=5*2 = 10

answer is option C
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If x and y are positive integers and 1620x/y^2 is the square of an odd 15 Sep 2020, 05:26
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energetics
If x and y are positive integers and \(\frac{1620x}{y^2}\) is the square of an odd integer, what is the smallest possible value of xy?

A) 1
B) 8
C) 10
D) 15
E) 28

Show more
Solution:

If we let 1620x/y^2 = z^2 (where z is odd), then 1620x = y^2 * z^2. Since the product of two perfect squares is also a perfect square, 1620x must itself be a perfect square.

Since 1620 = 81 * 20 = 3^4 * 2^2 * 5, we see that x must be at least 5 so that 1620x is a perfect square. Since 1620x/y^2 is the square of an odd integer, we see that y must be at least 2, so that if x = 5, we have 1620x/y^2 = (3^4 * 2^2 * 5^2) / (2^2) = 3^4 * 5^2 = (3^2 * 5)^2.

Therefore, the smallest possible value of xy is 5 * 2 = 10.

Answer: C
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If x and y are positive integers and 1620x/y^2 is the square of an odd 22 Oct 2025, 03:45
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