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?

Solution 1

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

Solution 2

Let's divide all elements of T into sections. Each section ranges from $10^{i}$ to $10^{i+1}$ And, each section must have 1 or 2 elements. So, let's consider both cases.

If a section has 1 element, we claim that the number doesn't have 9 as the leftmost digit. Let this element be $9^{k}$ and the section ranges from $10^{i}$ to $10^{i+1}$. To the contrary, let's assume the number ($9^{k}$) does have 9 as the leftmost digit. Thus, $9 \cdot 10^i \leq 9^k$. But, if you divide both sides by 9, you get $10^i \leq 9^{k-1}$, and because $9^{k-1} < 9^{k} < 10^{i+1}$, so we have another number ($9^{k-1}$) in the same section ($10^i \leq 9^{k-1} < 10^{i+1}$). Which is a contradiction to our assumption that the section only has 1 element. So in this case, the number doesn't have 9 as the leftmost digit.

If a section has 2 elements, we claim one has to have a 9 as the leftmost digit, one doesn't. Let the elements be $9^{k}$ and $9^{k+1}$, and the section ranges from $10^{i}$ to $10^{i+1}$. So We know $10^{i} \leq 9^{k} < 9^{k+1} < 10^{i+1}$. From $10^{i} \leq 9^{k}$, we know $9 \cdot 10^{i} \leq 9^{k+1}$, and since $9^{k+1} < 10^{i+1}$. The number($9^{k+1}$)'s leftmost digit must be 9, and the other number($9^{k}$)'s leftmost digit is 1.

There are total 4001 elements in T and 3817 sections that have 1 or 2 elements. And, no matter how many elements a section has, each section contains exactly one element that doesn't begin with 9. We can take 4001 elements, subtract 3817 elements that don't have 9 as the leftmost digit, and get $\boxed{184}$ numbers that have 9 as the leftmost digit.

- AlexLikeMath