Grades 11-12 Video Solutions 2012-2012_11-12

Video Transcription

Consider two operations which can be performed on a fraction. First is to increase the numerator by 8, and second is to increase the denominator by 7.  When we start with the fraction 7 8ths and perform a total of n such operations in some order, we obtain 7 8ths again. What's the smallest possible value of n?  So let's say we do x number of the first operation and y of the second operation. That means we want to solve the equation 7 plus 8x over 8 plus 7y equals 7 8ths. We want to find  the smallest values of x and y. We want to find the smallest value of x plus y, because that's  the total of number of moves where this can happen. So let's cross multiply and then cancel  out these 7 times 8 that show up on both sides, and now we're here. This side is divisible by 49,  and so this side must also be divisible by 49, and so that means x is a multiple of 49. Similarly, y has to be a multiple of 64. So since we want to minimize x plus y, we see that  x equals 49 and y equals 64 will do the trick for us. It satisfies this equation, and so x plus y equals 113 does the trick.

Video Summary

The problem involves finding the smallest number of operations needed to return a fraction to its original form after altering the numerator and denominator. Starting with the fraction 7/8, the operations are: increasing the numerator by 8, and increasing the denominator by 7. To solve 7 + 8x over 8 + 7y = 7/8 with minimal operations, we determine that x must be a multiple of 49 and y a multiple of 64. The smallest solution for x and y, thus minimizing x + y, is x = 49 and y = 64, resulting in a total of 113 operations.

Keywords

fraction

operations

numerator

denominator

solution