12. Let $n$ be a positive integer such that the decimal representation of $2^{n}$ starts with the digit 7 (i.e., the leftmost non-zero digit is 7). With which digit does the decimal representation of $5^{n}$ start?
(A) 1
(B) 2
(C) 5
(D) 7
(E) It depends on $n$

## Numerical Answer Problems - 5 points
12. The answer is (A). It can be found by exploiting the equality $5^{n}=\frac{10^{n}}{2^{n}}$. Let $c$ be the number of decimal digits of $2^{n}$; the long division algorithm implies that the first digit of the quotient $\frac{10^{n}}{2^{n}}$ is the result of the integer division of $10^{c}$ by $2^{n}$. This result is 1, since $2^{n} \geq 7 \cdot 10^{c-1}$ and thus $2 \cdot 2^{n} \geq 14 \cdot 10^{c-1} > 10^{c}$.

Alternatively, it is possible to express the solution formally without referring to the long division algorithm. Let $x$ be the initial digit of $5^{n}$; there exist integers $k, h$ such that $7 \cdot 10^{k} \leq 2^{n} < 8 \cdot 10^{k}$ and $x \cdot 10^{h} \leq 5^{n} < (x+1) \cdot 10^{h}$. Multiplying these inequalities term by term and using the fact that $2^{n} \cdot 5^{n} = 10^{n}$, we obtain

$$
7 x \cdot 10^{k+h} \leq 10^{n} < 8(x+1) \cdot 10^{k+h}
$$

from which in particular $10^{k+h} < 10^{n} < 10^{k+h+2}$. Clearly, this is only possible if $n = k + h + 1$. From the inequality $7 x \cdot 10^{k+h} < 10^{n} = 10^{k+h+1}$, we then get $7 x < 10$, which (since $x$ is a positive integer) is only possible for $x = 1$.