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If x and y are positive integers and xy=36, what is the smal

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by Max@Math Revolution » Mon Aug 27, 2018 1:37 am

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[Math Revolution GMAT math practice question]

If x and y are positive integers and xy=36, what is the smallest possible value of x+y?

A. 6
B. 8
C. 10
D. 12
E. 18
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by fskilnik@GMATH » Mon Aug 27, 2018 4:57 am
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by Brent@GMATPrepNow » Mon Aug 27, 2018 5:40 am
Max@Math Revolution wrote:[Math Revolution GMAT math practice question]

If x and y are positive integers and xy=36, what is the smallest possible value of x+y?

A. 6
B. 8
C. 10
D. 12
E. 18
Since there aren't many PAIRS of integers that have a product of 36, we can simply list the possibilities:

x = 1 and y = 36. Here, the SUM = 37
x = 2 and y = 18. Here, the SUM = 20
x = 3 and y = 12. Here, the SUM = 15
x = 4 and y = 9. Here, the SUM = 13
x = 6 and y = 6. Here, the SUM = 12

Answer: E

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Brent
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by Max@Math Revolution » Wed Aug 29, 2018 12:50 am
=>

When we know the product of two numbers, the sum of those numbers is a minimum when the two numbers are equal.

Since xy = 36, the minimum value of x + y occurs when x = y = 6. It is 6 + 6 = 12.

Therefore, D is the answer.
Answer: D
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by fskilnik@GMATH » Wed Aug 29, 2018 4:52 am
Max@Math Revolution wrote: When we know the product of two numbers, the sum of those numbers is a minimum when the two numbers are equal.
This is an immediate consequence of the famous Arithmetic-Geometric Mean inequality , therefore out-of-scope of the GMAT.

Important detail: the condition about integers (given in the question stem) is unnecessary, but the condition on positive (in fact nonnegative) numbers is essential.

For the interested readers:
[Arithmetic-Geometric Mean inequality] wrote: Given any positive integer \[N \geqslant 2\] and \[{x_1}\,,\,\,{x_2}\,,\, \ldots ,\,\,{x_N}\] nonnegative real numbers , we have:
\[\sqrt[{N\,}]{{{x_1}\, \cdot \,\,{x_2} \cdot \, \ldots \cdot \,\,{x_N}}}\,\,\, \leqslant \,\,\,\frac{{{x_1}\, + \,\,{x_2}\, + \, \ldots + \,{x_N}}}{N}\]
with equality if, and only if, all real numbers considered are equal.
In our case, \[N = 2\] and we have \[\sqrt {36} = \sqrt {xy} \leqslant \,\,\frac{{x + y}}{2}\] and (x+y) will be minimum exactly when (x+y)/2 is minimum...

This occurs (by the famous theorem above) only when x and y are equal, i.e., when \[6 = \sqrt {36} = \sqrt {xy} = \sqrt {{x^2}} \,\,\,\mathop = \limits^{x\, \geqslant \,0} \,x\,\,\,\,\left( { = y} \right)\]

Therefore \[? = 6 + 6\] as mentioned.

I hope you find all this interesting.

Regards,
fskilnik.
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by Jeff@TargetTestPrep » Wed Sep 05, 2018 9:44 am
Max@Math Revolution wrote:[Math Revolution GMAT math practice question]

If x and y are positive integers and xy=36, what is the smallest possible value of x+y?

A. 6
B. 8
C. 10
D. 12
E. 18
The possible combinations of x and y are:

1, 36

2, 18

3, 12

4, 9

6, 6

So the smallest possible value of x + y is 12.

Alternate Solution:

It is possible to interpret the problem in the following way: Among all the rectangles that have an area of 36, what is the smallest possible sum for the length and width? We know that among all rectangles that have the same area, the square has the smallest perimeter. If the perimeter is the smallest, then so will be the sum of length and width (since perimeter of a rectangle is simply twice the sum of length and width). Thus, the smallest possible value is 6 + 6 = 12.

Answer: D

Jeffrey Miller
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