Show that there exists an integer that is a power of 5 with the first 4 digits being 2018.

Question

I know the problem can be expressed in the following way. Let $a,b \in \mathbb{Z}^+$ such that $2018\:\times\:10^b < 5^a < 2019\:\times\:10^b$

Taking logs, I get $b + \log 2018 < a \log 5< b + \log 2019$.

Not quite sure how else to proceed though.

Answer 1

The set $A = \{ a \log_{10} 5 + b : a,b \in \mathbb Z \}$ is an additive subgroup of $\mathbb R$, and so is either cyclic or dense.

If $A$ were cyclic, then $1 = n (a_0 \log_{10} 5 + b_0)$. Since $\log_{10} 5$ is irrational, we must have $n a_0 =0$ and $n b_0 = 1$, which cannot happen.

Therefore, $A$ is dense and so there is an element of $A$ in the interval $(\log_{10}(2018), \log_{10}(2019))$.

Answer 2

Hint Since $\log_{10}(5)$ is irrational, the set $\{ a \log_{10}(5)-b : a, b \in \mathbb N\}$ is dense in $\mathbb R$. Therefore, you can find such a number in the interval $(\log_{10}(2018), \log_{10}(2019))$.