If and are positive integers such that when divided by , leaves a remainder 29, what is the minimum possible value of ?
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
Since x on division by y leaves remainder 2 9 , we can write this situation mathematically as x = k y + 2 9 where k is a non-negative number. We have to find the minimum value of x y , which we can write as- x y = ( k y + 2 9 ) y = k y 2 + 2 9 y Now, to minimize the above reached expression, we must minimize k and y . Since k is a non-negative number, it's minimum value is 0 . Also, we know that if a number (say, a ) leaves a remainder r on division by b , that is, a = b q + r , then b > r . Therefore, in this case, y > 2 9 . So the minimum possible integer value of y is 3 0 . Now that we have all the values, we can put them in the expression to get the minimum value as- 0 × 3 0 2 + 2 9 ⋅ 3 0 = 2 9 × 3 0 = 8 7 0 And we are done.