A 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 \(3^m \times 5^n\) (where \(m\) and \(n\) are integers) have exactly \(10\) proper factors.
Any factor of \(3^2 \times 5^3\) must be of the form \(3^r \times 5^s\) with \(0 \leq r \leq 2\) and \(0 \leq s \leq 3\).
This gives \(3\) possible values for \(r\) and \(4\) for \(s\), so there are \(3 \times 4 = 12\) factors.
But two of these are not proper factors (namely \(1\) and \(3^2 \times 5^3\) itself), so there are \(12 - 2 = 10\) proper factors.
We can list all twelve factors as
\(3^0 \times 5^0 \mathbin{(=} 1)\) (not a proper factor)
\(3^0 \times 5^1 \mathbin{(=} 5)\)
\(3^0 \times 5^2 \mathbin{(=} 25)\)
\(3^0 \times 5^3 \mathbin{(=} 125)\)
\(3^1 \times 5^0 \mathbin{(=} 3)\)
\(3^1 \times 5^1 \mathbin{(=} 15)\)
\(3^1 \times 5^2 \mathbin{(=} 75)\)
\(3^1 \times 5^3 \mathbin{(=} 375)\)
\(3^2 \times 5^0 \mathbin{(=} 9)\)
\(3^2 \times 5^1 \mathbin{(=} 45)\)
\(3^2 \times 5^2 \mathbin{(=} 225)\)
\(3^2 \times 5^3 \mathbin{(=} 1125)\) (not a proper factor).
Now that we understand how to count the factors of \(3^2 \times 5^3\), we can answer the second part.
The number of proper factors of \(3^m \times 5^n\) is \((m + 1)(n + 1) - 2\), as the power of \(3\) in a factor can be \(0\), \(1\), …, or \(m\), and the power of \(5\) can be \(0\), \(1\), …, or \(n\).
So we require \((m + 1)(n + 1) - 2 = 10\), or equivalently \((m + 1)(n + 1) = 12\).
\(12\) can be factorised into 2 factors (where order matters) in the 6 ways
\(12=1\times12\),
\(12=2\times6\),
\(12=3\times4\),
\(12=4\times3\),
\(12=6\times2\) and
\(12=12\times1\),
which give exactly six possible integers of the form \(3^m \times 5^n\): \[3^0 \times 5^{11}, 3^1 \times 5^5, 3^2 \times 5^3, 3^3 \times 5^2, 3^5 \times 5^1, 3^{11} \times 5^0.\]
So there are \(6\) possibilities in total. This means that there are \(5\) other integers with the required properties.
- Let \(N\) be the smallest positive integer that has exactly \(426\) proper factors. Determine \(N\), giving your answer in terms of its prime factors.
Let \(n=2^a \times 3^b \times 5^c \times 7^d \times ...\). The number of proper factors of \(n\) is \((a+1)(b+1)(c+1)(d+1)...-2\) and we require this to be equal to \(426\).
That is, \[(a+1)(b+1)(c+1)(d+1)...=428.\]
The prime factorisation of \(428\) is \(428=2^2 \times 107\).
Therefore all the possible factorisations of \(428\) into two or more factors are
\(428=1 \times 428\),
\(428=2 \times 214\),
\(428=2 \times 2 \times 107\) and
\(428=4 \times 107\).
We want to find the smallest possible \(N\), so we must use the smallest primes and of these we must give the smallest primes the largest powers and the largest primes the smallest powers.
Therefore we know that the smallest \(N\) must be one of
\(2^{427}\),
\(2^{213} \times 3\),
\(2^{106} \times 3 \times 5\) and
\(2^{106} \times 3^3\).
The smallest of these is \(2^{106} \times 3 \times 5\) because if we divide each solution by \(2^{106}\) we obtain
\(2^{321}\),
\(2^{107} \times 3\),
\(3 \times 5\) and
\(3^3\),
where \(3 \times 5\) is the smallest.
Therefore \[N=2^{106} \times 3 \times 5.\]
