There is a certain number N = n 1 ⋅ n 2 ⋅ n 3 ⋯ n k , where 1 ≤ n i ≤ 1 0 0 ( 1 ≤ i ≤ k ) are all distinct positive integers.
We are told the following:
If 2 0 1 6 ⋅ N has B A d divisors, with A and B coprime positive integers, submit A + B .
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Take n 1 = 2 8 , n 2 = 3 5 , n 3 = 4 0 , n 4 = 4 9 , n 5 = 5 0 , n 7 = 7 2 , n 8 = 9 8 .
I don't think you can do it with fewer factors!
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Relevant wiki: Number of Factors
Key fact: if the prime factorization of N is N = 2 e 2 ⋅ 3 e 3 ⋅ 5 e 5 ⋯ p e p ⋯ , then the number of divisors of N is δ ( N ) = ( 1 + e 2 ) ( 1 + e 3 ) ( 1 + e 5 ) ⋯ ( 1 + e p ) ⋯ .
From the fact that δ ( N ) is odd, we know that all exponents e i must be even, so that N is a perfect square. In particular, this means that N contains no prime factors greater than 50.
Using 1 2 8 = 2 7 , we have δ ( N ) δ ( 1 2 8 N ) = ( 1 + e 2 ) ( 1 + e 3 ) ⋯ ( 1 + e 2 + 7 ) ( 1 + e 3 ) ⋯ = 1 + e 2 8 + e 2 = 1 1 1 8 , which tells us immediately that e 2 = 1 0 .
Now 2 1 3 = 3 ⋅ 7 1 , so δ ( N ) δ ( 2 1 3 N ) = ( 1 + e 3 ) ( 1 + e 7 1 ) ( 2 + e 3 ) ( 2 + e 7 1 ) = 3 8 . We already know that e 7 1 = 0 , so that we find 1 + e 3 2 + e 3 = 3 4 , making e 3 = 2 .
Next, 1 7 5 = 7 ⋅ 5 2 , giving δ ( N ) δ ( 1 7 5 N ) = ( 1 + e 5 ) ( 1 + e 7 ) ( 3 + e 5 ) ( 2 + e 7 ) = 5 8 . To solve this, we cross-multiply, work out brackets, and subtract, and refactor (I find it convenient to multiply everything by 3): 5 ( 3 + e 5 ) ( 2 + e 7 ) = 8 ( 1 + e 5 ) ( 1 + e 7 ) ; 3 0 + 1 0 e 5 + 1 5 e 7 + 5 e 5 e 7 = 8 + 8 e 5 + 8 e 7 + 8 e 5 e 7 ; ( 3 e 5 − 7 ) ( 3 e 7 − 2 ) = 8 0 ; there are ten positive integer factorizations of 80, and only four of them result in integer values for e 5 and e 7 . Of these four, only one has even values: e 5 = 4 and e 7 = 6 .
Thus we know that N = 2 1 0 ⋅ 3 2 ⋅ 5 4 ⋅ 7 6 ⋯ Using 2 0 1 6 = 2 5 ⋅ 3 2 ⋅ 7 , we get 2 0 1 6 N = 2 1 5 ⋅ 3 4 ⋅ 5 4 ⋅ 7 7 ⋯ and B A = δ ( N ) δ ( 2 0 1 6 N ) = 1 1 ⋅ 3 ⋅ 5 ⋅ 7 1 6 ⋅ 5 ⋅ 5 ⋅ 8 = 2 3 1 6 4 0 , making the answer 8 7 1 .