If the product of all the factors of a positive integer, N, is 2 ^ 18 * 3 ^ 12, how many values can N take? Answers.
A) None B) 1 C) 2 D) 3 E) 4
Lets say   2^0 *3^0 * 2^0 * 3^1....2^0 * 3^n ...A0 &gt;&gt; 2^{0*(N+1)} * 3^{N(N+1)/2}   2^1 *3^0 * 2^1 * 3^1....2^1 * 3^n ...A2 &gt;&gt; 2^{1*(N+1)} * 3^{N(N+1)/2}   .   .   .   2^m *3^0 * 2^m * 3^1....2^m * 3^n ...AM &gt;&gt; 2^{M*(N+1)} * 3^[N(N+1)/2}   So product of all the factors are: A0 * A1 *...*AM = 2^{M(M+1)*(N+1)/2} * 3^{N(N+1)*(M+1)/2}   From the equation:   M(M+1)*(N+1)/2 = 18 --- X   N(N+1)*(M+1)/2 = 12 --- Y   dividing X, Y, we get M/N = 3/2   Lets say M = 3X, N = 2X   substituting the value in X OR Y we got X = 1 is the only integer satisfying   Hence the number is 2 ^ 3 * 3 ^ 2   Answer is 1
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If the product of all the factors of a positive integer, N, is 2 ^ 18 * 3 ^ 12, how many values can N take?