2018 AMC 10A Problems/Problem 22
Let and be positive integers such that , , , and . Which of the following must be a divisor of ?
We can say that and 'have' , that and have , and that and have . Combining and yields has (at a minimum) , and thus has (and no more powers of because otherwise would be different). In addition, has , and thus has (similar to , we see that cannot have any other powers of ). We now assume the simplest scenario, where and . According to this base case, we have . We want an extra factor between the two such that this number is between and , and this new factor cannot be divisible by or . Checking through, we see that is the only one that works. Therefore the answer is
Solution by JohnHankock
Elaborating on to what Solution 1 stated, we are not able to add any extra factor of 2 or 3 to because doing so would later the of and . This is why:
The is and the of is . However, the of (meaning both are divisible by 36). Therefore, is only divisible by (and no higher power of 3), while is divisible by only (and no higher power of 2).
Thus, the of can be expressed in the form for which is a number not divisible by or . The only answer choice that satisfies this (and the other condition) is .
Solution 2 (Better notation)
First off, note that , , and are all of the form . The prime factorizations are , and , respectively. Now, let and be the number of times and go into ,respectively. Define , , , and similiarly. Now, translate the s into the following: .
(Unfinished) ~Rowechen Zhong
|2018 AMC 10A (Problems • Answer Key • Resources)|
|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 10 Problems and Solutions|