2016 AMC 12B Problems/Problem 24
Problem
There are exactly ordered quadruplets such that and . What is the smallest possible value for ?
Solution
Let , etc., so that . Then for each prime power in the prime factorization of , at least one of the prime factorizations of has , at least one has , and all must have with .
Let be the number of ordered quadruplets of integers such that for all , the largest is , and the smallest is . Then for the prime factorization we must have So let's take a look at the function by counting the quadruplets we just mentioned..
There are quadruplets which consist only of and . Then there are quadruplets which include three different values, and with four. Thus and the first few values from onwards are Straight away we notice that , so the prime factorization of can use the exponents . To make it as small as possible, assign the larger exponents to smaller primes. The result is , so which is answer .
See Also
2016 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 23 |
Followed by Problem 25 |
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 |