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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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Senior Manager
Status: Gathering chakra
Joined: 05 Feb 2018
Posts: 443
If x and y are positive integers and 1620x/y^2 is the square of an odd  [#permalink]

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19 Apr 2019, 11:23
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35% (medium)

Question Stats:

74% (02:16) correct 26% (02:16) wrong based on 90 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:
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
Intern
Joined: 09 Apr 2019
Posts: 8
Location: Brazil
Schools: Wharton, Kellogg, Booth
GPA: 3.45
If x and y are positive integers and 1620x/y^2 is the square of an odd  [#permalink]

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19 Apr 2019, 15:28
5
$$\frac{1620x}{y^2} = z^2$$, 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 $$xy = 10$$

##### General Discussion
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Joined: 24 Nov 2016
Posts: 1218
Location: United States
If x and y are positive integers and 1620x/y^2 is the square of an odd  [#permalink]

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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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Joined: 20 Dec 2019
Posts: 35
Re: If x and y are positive integers and 1620x/y^2 is the square of an odd  [#permalink]

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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
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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