1990 AIME Problems/Problem 13

Revision as of 01:18, 10 February 2023 by Mathandski (talk | contribs)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)


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

Lemma: For all positive integers n, there's exactly one n-digit power of 9 that does not have a left-most digit 9

(Not-so-rigorous) Proof: One can prove by contradiction that there must be at least either one or two n-digit power of 9 for all n. If there is exactly 1 n-digit power of 9, then such a number $m$ cannot begin with 9 since that would result in $\frac{m}{9}$ also being an n-digits, hence a contradiction. Therefore, this single n-digit power of 9 must not begin with 9. In the case that there are 2 n-digit powers of 9, we have already discovered that one of them must begin with 9, let it be $9^k = m$. The integer $9^{k - 1} = \frac{m}{9}$ must then be an n-digit number that begins with $1$.

Using this lemma, out of the 4001 powers of 9 in T, exactly 3816 must not begin with 9 (one for each digit). Therefore, the answer is $4000 - 3816 = \boxed{184}$ numbers have 9 as their leftmost digits.

~Edited by Mathandski (with help from a math god)

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

Solution 3

We know that $9$ is very close to $10$. But we also know that for all $k \in \mathbb{Z}^+$, $10^k$ has $k+1$ digits. Hence $10^{4000}$ has $4001$ digits. Now, notice that if a power of $9^n$ has $9$ as the leftmost digit, then $9^{n-1}$ has $1$ as the leftmost digit and will preserve the same number of digits (just due to division). Hence we find that consecutive powers of $9$ are "supposed" to increase by one digit, however, when consecutive powers of $9$ have leading digit $1$ then leading digit $9$ they "lose" this one digit. Therefore, it suffices to find the number of times that $9^k$ has "lost" a digit since the number of times that this occurs shows us the number of pairs where the leading digit of $9^k$ is $1$ and the leading digit of $9^{k+1}$ is $9$. Hence, we find that $9^{4000}$ is "supposed" to have $4001$ digits, but only has $3817$ digits. Thus, the number of times $9^k$ has lost a digit is $4001-3817 = 184$. Thus, $\boxed{184}$ elements of $T$ have $9$ as their leading digit.


See also

1990 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png