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Smallest possible value of xy

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by Troika » Tue May 29, 2012 1:22 am
If x and y are positive integers and x/y has a remainder of 5, what is the smallest possible value of xy?

OA: 30

Source: MGMAT NP, pg. 134
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by Anurag@Gurome » Tue May 29, 2012 1:38 am
HG10 wrote:If x and y are positive integers and x/y has a remainder of 5, what is the smallest possible value of xy?
As x and y are positive integers, to minimize xy, we have to minimize both x and y.

When x is divided by y, the remainder is 5.
Hence, y must be greater than 5.
Therefore, minimum possible value of y is 6.

Now, possible values of x such that when divided by 6 we get a remainder of 5 are 5, 11, 17 etc. Hence, minimum possible value of x is 5.

Hence, minimum possible value of xy = 5*6 = 30
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by Anurag@Gurome » Tue May 29, 2012 1:55 am
Algebraic Approach:
x can be written as (ky + 5), where k is any non-negative integer and y > 5.

Hence, xy = (ky + 5)*y = (ky² + 5y)
To minimize (ky² + 5y), we have to minimize both k and y.

As k is a non-negative integer, minimum possible value of k is 0 and minimum possible value of y is 6.

Hence, minimum possible value of xy is (0*6 + 5*6) = 30
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by Troika » Tue May 29, 2012 2:04 am
Anurag, in your first post I don't understand how 5 divided by 6 could give a remainder of 5. I understand 11, 17 etc. Please explain. Thanks
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by usfall13ivy » Tue May 29, 2012 2:17 am
HG10 wrote:Anurag, in your first post I don't understand how 5 divided by 6 could give a remainder of 5. I understand 11, 17 etc. Please explain. Thanks
The idea behind this is that if you divide 5by 6, you would get, 0.834 which is a real number and hence it is considered that 5 is indeed divisible by 6. But when you start the division, the quotient is 0 and the remainder left is 5, its in the next step we add a decimal point and start with the division and get .834. Thats why the first number to be divided by 6 that would leave a remainder of 5 is 5.
Hope that helped.
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by LalaB » Tue May 29, 2012 4:37 am
5/6 HAS A REMAINDER 5

so, 5*6 =30
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