Why does $10^x$ have $(x+1)^2$ factors?

eg. $1000$ has $16$ factors $(1, 2, 4, 5, 8, 10, 20, 25, 40, 50, 100, 125, 200, 250, 500, 1000)$

Solution 1:

Assuming $x$ is a positive integer, of course...

How many factors does a positive integer $n$ have? If you factor $n$ into primes, $$n = p_1^{a_1}\times p_2^{a_2}\times\cdots\times p_r^{a_r}$$ then every factor of $n$ is of the form $$m = p_1^{b_1}\times p_2^{b_2}\times\cdots\times p_r^{b_r}$$ with $0\leq b_i\leq a_i$. So the number of factors of $n$ is $$(a_1+1)(a_2+1)\cdots(a_r+1)$$ (because you have $a_1+1$ choices for the exponent of $p_1$, $(a_2+1)$ choices for the exponent of $p_2$, etc.)

Since $10 = 2\times 5$, then $10^x = 2^x\times 5^x$, so the number of factors is $(x+1)(x+1)=(x+1)^2$.

Solution 2:

The short answer to your question is $10^x = 2^x 5^x$. We shall now enumerate the number of divisors. Let $d$ be one such divisor. Then $d = 2^y 5^z$, where $0 \leq y \leq x$ and $0 \leq z \leq x$. Now $y$ has $x+1$ choices and similarly $z$ has $x+1$ choices and each of these choices are independent and will give rise to an unique divisor $d$.

Hence, the total number of divisors of $10^x$ is $(x+1) \times (x+1) = (x+1)^2$.

In general, to find the number of divisors of $n$, write $n = p_1^{\alpha_1} \times p_2^{\alpha_2} \times \ldots p_k^{\alpha_k}$.

A divisor $d$ of $n$ must be of the form $d = p_1^{\beta_1} \times p_2^{\beta_2} \times \ldots p_k^{\beta_k}$ where $0 \leq \beta_i \leq \alpha_i$, $i \in \{0,1,2,\ldots,k\}$. Hence, each $\beta_i$ has $(1+\alpha_i)$ choices and each of these choices are independent and will give rise to an unique divisor $d$.

Hence, the total number of divisors of $n$ is $$\displaystyle \prod_{i=1}^{k} (1+\alpha_i)$$

Solution 3:

${\displaystyle 10^x = 2^x5^x}$. A factor of $10^x$ is a number of the form ${\displaystyle 2^y5^z}$ where $0 \leq y \leq x$ and $0 \leq z \leq x$, $y$ and $z$ integers. There are $(x+1)$ integers from $0$ to $x$, so the total number of possible $(y,z)$ is $(x+1)*(x+1) = (x+1)^2$.