2003 AMC 12B Problems/Problem 18
Let and be positive integers such that The minimum possible value of has a prime factorization What is
Substitute into . We then have . Divide both sides by , and it follows that:
Note that because and are prime, the minimum value of must involve factors of and only. Thus, we try to look for the lowest power of such that , so that we can take to the fifth root. Similarly, we want to look for the lowest power of such that . Again, this allows us to take the fifth root of . Obviously, we want to add to and subtract from because and are multiplied by and divided by , respectively. With these conditions satisfied, we can simply multiply and and substitute this quantity into to attain our answer.
We can simply look for suitable values for and . We find that the lowest , in this case, would be because . Moreover, the lowest should be because . Hence, we can substitute the quantity into . Doing so gets us:
Taking the fifth root of both sides, we are left with .
A simpler way to tackle this problem without all that modding is to keep the equation as:
As stated above, and must be the factors 7 and 11 in order to keep at a minimum. Moving all the non-y terms to the left hand side of the equation, we end up with:
The above equation means that must also contain only the factors 7 and 11 (again, in order to keep at a minimum), so we end up with:
( and are arbitrary variables placed in order to show that could have more than just one 7 or one 11 as factors)
Since 7 and 11 are prime, we know that and . The smallest positive combinations that would work are and . Therefore, . is correct.
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