2003 AMC 12B Problems/Problem 18
Problem
Let 
and 
be positive integers such that 
The minimum possible value of has a prime factorization 
What is 
Solution
Substitute 
into . We then have 
. Divide both sides by 
, and it follows that:
(a^{5c}b^{5d}) = \[\frac{11y^{13}}{7}. (Error compiling LaTeX. Unknown error_msg)
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:
(a^{5c}b^{5d}) = \[\frac{11(11^{3} \cdot 7^{2})^{13}}{7} = 11^{40} \cdot 7^{25}. (Error compiling LaTeX. Unknown error_msg)
Taking the fifth root of both sides, we are left with 
. 
See Also
| 2003 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 17 |
Followed by Problem 19 |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
| All AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
