Divisor Product Puzzle
Given that $ n > 1 $, what is the smallest positive integer $ n $ whose positive divisors have a product of $ n^6 $?
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- $\frac{a}{b}$
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- $a^n$
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- $a_n$
- $\sqrt{}$
- $\sqrt[n]{}$
- $\pi$
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- $\infty$
Solution
Let's multiply the divisors of a positive integer, say $ 12 $. The divisors of $ 12 $ are $ 1,2,3,4,6, $ and $ 12 $. The product of the divisors of 12 is $ 1\cdot2\cdot3\cdot4\cdot6\cdot12=(1\cdot12)(2\cdot 6)(3\cdot4)=12^3 $. The factors may be regrouped in this way for any positive integer with an even number of divisors. We have found that if the number $ d $ of divisors is even, then the product of the divisors of $ n $ is $ n^{d/2} $. Solving $ n^6=n^{d/2} $, we find $ d=12 $.
Recall that we can determine the number of factors of $ n $ by adding $ 1 $ to each of the exponents in the prime factorization of $ n $ and multiplying the results. We work backwards to find the smallest positive integer with $ 12 $ factors. Twelve may be written as a product of integers greater than 1 in four ways: $ 12 $, $ 2\cdot 6 $, $ 3\cdot 4 $, and $ 2\cdot2\cdot3 $. The prime factorizations which give rise to these products have sets of exponents $ \{11\} $, $ \{5,1\} $, $ \{3,2\} $, and $ \{2,1,1\} $. In each case, we minimize $ n $ by assigning the exponents in decreasing order to the primes $ 2,3,5,\ldots $. Therefore, the smallest positive integer with 12 factors must be in the list $ 2^{11}=2048 $, $ 2^5\cdot3=96 $, $ {2^3\cdot3^2}=72 $, and $ 2^2\cdot3\cdot5=60 $. The smallest of these is $ \boxed{60} $.