A {\em proper factor} of an integer \(N\) is a positive integer, not \(1\) or \(N\), that divides \(N\).
Show that \(3^2\times 5^3\) has exactly \(10\) proper factors. Determine how many other integers of the form \(3^m\times5^n\) (where \(m\) and \(n\) are integers) have exactly 10 proper factors.
Let \(N\) be the smallest positive integer that has exactly \(426\) proper factors. Determine \(N\), giving your answer in terms of its prime factors.
Solution:
All factors of \(3^2 \times 5^3\) have factors of the form \(3^k \times 5^l\) where \(0 \leq k \leq 2\) and \(0 \leq l \leq 3\) therefore there are \(3\) possible values for \(k\) and \(4\) possible values for \(l\), which gives \(3 \times 4 = 12\) factors, which includes \(2\) factors we aren't counting, so \(10\) proper factors.
By the same argument \(3^m \times 5^n\) has \((m+1) \times (n+1) - 2\) proper factors, so we want \((m+1) \times (n+1) = 12\), so we could have
\begin{array}{cccc}
\text{factor} & m+1 & n + 1 & m & n \\
12 = 12 \times 1 & 12 & 1 & 11 & 0 \\
12 = 6 \times 2 & 6& 2 & 5 & 1 \\
12 = 4 \times 3 & 4& 3 & 3 & 2 \\
12 = 3 \times 4 & 3& 4 & 2 & 3 \\
12 = 2 \times 6 & 2& 6 & 1 & 5 \\
12 = 1 \times 12 & 1& 12 & 0 & 11 \\
\end{array}
So we could have \(3^{11}, 3^{5} \times 5^1 3^3 \times 5^2, 3^2 \times 5^3, 3^1 \times 5^5, 5^{11}\)
Suppose \(N\) has \(426\) proper factors, then it has \(428 = 2^2 \times 107\) factors, so it will either factor as \(p^{427}\) or \(p_1^{106} p_2^{3}\) or \(p_1^{106} p_2 p_3\). Clearly the first will be very large, and we should have \(p_1 < p_2 < p_3\), so lets consider \(2^{106}\) with either \(3^3 = 27\) or \(3 \times = 15 < 27\). Therefore we should take \(2^{106} \times 3 \times 5\)