# 1990 AIME Problems/Problem 13

## Problem

Let $T = \{9^k : k ~ \mbox{is an integer}, 0 \le k \le 4000\}$. Given that $9^{4000}_{}$ has 3817 digits and that its first (leftmost) digit is 9, how many elements of $T_{}^{}$ have 9 as their leftmost digit?

.

Since $9^{4000}$ has 3816 digits more than $9^1$, $4000 - 3816 = \boxed{184}$ numbers have 9 as their leftmost digits.
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