Number of positive divisors of a number

Question

Prove that the number of positive divisors of 1992 1s: $111...1$ is even.

I have no idea tackling this question. Can anyone provide any hint? Thanks.

Answer 1

Notice that if $k$ is a divisor of $n$, $\frac{n}{k}$ is also a divisor of $n$. In other words, the positive divisors of a number come in pairs. That means all numbers have an even number of positive divisors unless $k=\frac{n}{k}$. Because if $k$ does equal $\frac{n}{k}$, you discard one of them from your count. You don't count the same thing twice.

If you have understood everything up to this point, you should realize that a number can have an odd number of positive divisors if and only if it is a perfect square.

Now you just have to show that your number is not a perfect square and you're done!

This is not difficult to do, because any perfect square is $\equiv 0, 1 \pmod 4$. And your number is $\equiv 3\pmod 4$. So it can't be a perfect square.

So it has to have an even number of positive divisors.

Answer 2

Hint: The number of divisors of primes is even (equals to $2$). The number of divisors of $4$ is odd ($\{1,2,4\}$). The number of divisors of $6$ is even ($\{1,2,3,6\}$). The number of divisors of $8$ is even ($\{1,2,4,8\}$). The number of divisors of $9$ is odd. The number of divisors of $10,12,14,15$ is even. The number of divisors of $16$ is odd.

Can you see a pattern?