How many divisors

Let n n be a positive integer which has 30 divisors inclusive of 1 and itself, find the minimum value for n n .


The answer is 720.

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1 solution

Matteo Monzali
Jun 3, 2016

The number of divisors of a number n n in the form n = α 1 p 1 α 2 p 2 . . . α n p n n=\alpha_1 ^{p_1} *\alpha_2 ^{p_2} *...*\alpha_n ^{p_n} where α 1 , α 2 . . . α n \alpha _1, \alpha _2 ... \alpha _n are prime numbers, is given by N d = ( p 1 + 1 ) ( p 2 + 2 ) . . . ( p n + 1 ) N_d=(p_1+1)*(p_2+2)*...*(p_n+1) .

The way to write 30 minimizing the value of n n is 30 = 5 3 2 n = 2 4 3 2 5 = 720 30=5*3*2 \rightarrow n=2^4*3^2*5=720

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