Difference between revisions of "1990 AIME Problems/Problem 13"

Latest revision as of 00:18, 10 February 2023

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

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.

~qwertysri987

@@ Line 3: / Line 3: @@
 ==Solution 1==
-Since <math>9^{4000}</math> has 3816 digits more than <math>9^1</math>, <math>4000 - 3816 = \boxed{184}</math> numbers have 9 as their leftmost digits.
+'''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 <math>m</math> cannot begin with 9 since that would result in <math>\frac{m}{9}</math> 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 <math>9^k = m</math>. The integer <math>9^{k - 1} = \frac{m}{9}</math> must then be an n-digit number that begins with <math>1</math>. ''
+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 <math>4000 - 3816 = \boxed{184}</math> numbers have 9 as their leftmost digits.
+~Edited by Mathandski (with help from a math god)
 ==Solution 2==

1990 AIME (Problems • Answer Key • Resources)
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