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 .
Also, to get the above formula of , we can also use the complementary counting by doing , while the first term is for the four integers to independently have k+1 choices each, with the second term indicating to subtract all the possibilities for the four integers to have values between 0 and k-1, and similarly the third term indicating to subtract all the possibilities for the four integers to have values between 1 and k, in the end the fourth term meaning the make up for the values between 1 and k-1.
See Also
2016 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 23 |
Followed by Problem 25 |
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