Review question

# What's the smallest integer with 426 proper factors? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R7465

## Solution

A proper factor of an integer $N$ is a positive integer, not $1$ or $N$, that divides $N$.

1. 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.

1. 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 $N=2^{106} \times 3 \times 5.$