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

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