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If x and y are positive integers and 1620x/y^2 is the square of an odd

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If x and y are positive integers and 1620x/y^2 is the square of an odd  [#permalink]

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New post 19 Apr 2019, 11:23
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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:
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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If x and y are positive integers and 1620x/y^2 is the square of an odd  [#permalink]

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New post 19 Apr 2019, 15:28
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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  [#permalink]

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New post 17 Jan 2020, 07:07
energetics wrote:
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


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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Re: If x and y are positive integers and 1620x/y^2 is the square of an odd  [#permalink]

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New post 25 Jan 2020, 10:29
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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Re: If x and y are positive integers and 1620x/y^2 is the square of an odd   [#permalink] 25 Jan 2020, 10:29
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