Difference between revisions of "2004 AIME I Problems/Problem 15"
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f(x)=\begin{cases}1 & \text{if x = 1}}\\ \frac x{10} & \text{if x is divisible by 10}\\ x+1 & \text{otherwise}\end{cases} | f(x)=\begin{cases}1 & \text{if x = 1}}\\ \frac x{10} & \text{if x is divisible by 10}\\ x+1 & \text{otherwise}\end{cases} | ||
</cmath> | </cmath> | ||
− | and define a sequence as follows: <math>x_1=x</math> and <math>x_{n+1}=f(x_n)</math> for all positive integers <math>n</math>. Let <math>d(x)</math> be the smallest <math>n</math> such that <math>x_n=1</math>. (For example, <math>d(100)=3</math> and <math>d(87)=7</math>.) Let <math>m</math> be the number of positive integers <math>x</math> such that <math>d(x)=20</math>. Find the sum of the distinct prime factors of <math>m</math>. | + | and define a [[sequence]] as follows: <math>x_1=x</math> and <math>x_{n+1}=f(x_n)</math> for all positive integers <math>n</math>. Let <math>d(x)</math> be the smallest <math>n</math> such that <math>x_n=1</math>. (For example, <math>d(100)=3</math> and <math>d(87)=7</math>.) Let <math>m</math> be the number of positive integers <math>x</math> such that <math>d(x)=20</math>. Find the sum of the distinct prime factors of <math>m</math>. |
== Solution == | == Solution == | ||
− | {{ | + | We backcount the number of ways. Namely, we start at <math>x_{20} = 1</math>, which can only be reached if <math>x_{19} = 10</math>, and then we perform <math>18</math> operations that either consist of <math>A: (-1)</math> or <math>B: (\times 10)</math>. We represent these operations in a string format, starting with the operation that sends <math>f(x_{18}) = x_{19}</math> and so forth downwards. There are <math>2^9</math> ways to pick the first <math>9</math> operations; however, not all <math>9</math> of them may be <math>A: (-1)</math> otherwise we return back to <math>x_{10} = 1</math>, contradicting our assumption that <math>20</math> was the smallest value of <math>n</math>. By the [[complement principle]], we only have <math>2^9 - 1</math> ways. |
+ | |||
+ | Since we performed the operation <math>B: (\times 10)</math> at least once in the first <math>9</math> operations, it follows that <math>x_{10} \ge 20</math>, so that we no longer have to worry about reaching <math>1</math> again. Thus the remaining <math>9</math> operations can be picked in <math>2^9</math> ways, with a total of <math>2^9(2^9 - 1) = 2^{18} - 2^9</math> strings. | ||
+ | |||
+ | However, we must also account for a sequence of <math>10</math> or more <math>A: (-1)</math>s in a row, because that implies that at least one of those numbers was divisible by <math>10</math>, yet the <math>\frac{x}{10}</math> was never used, contradiction. We must use complement counting again by determining the number of strings of <math>A,B</math>s of length <math>18</math> such that there are <math>10</math> <math>A</math>s in a row. The first ten are not included since we already accounted for that scenario above, so our string of <math>10</math> <math>A</math>s must be preceded by a <math>B</math>. There are no other restrictions on the remaining seven characters. Letting <math>\square</math> to denote either of the functions, and <math>^{[k]}</math> to indicate that the character appears <math>k</math> times in a row, then our bad strings can take the forms: | ||
+ | <center><math>\begin{align*} | ||
+ | &\underbrace{BA^{[10]}}\square \square \square \square \square \square \square \square \\ | ||
+ | &\square \underbrace{BA^{[10]}}\square \square \square \square \square \square \square \\ | ||
+ | &\qquad \quad \cdots \quad \cdots \\ | ||
+ | &\square \square \square \square \square \square \square \underbrace{BA^{[10]}} \square \\ | ||
+ | &\square \square \square \square \square \square \square \square \underbrace{BA^{[10]}} \\ | ||
+ | \end{align*}</math></center> | ||
+ | |||
+ | There are <math>2^7</math> ways to select the operations for the <math>7</math> <math>\square</math>s, and <math>8</math> places to place our <math>BA^{[10]}</math> block. Thus, our answer is <math>2^{18} - 2^9 - 8 \cdot 2^7 = 2^9 \times 509</math>, and the answer is <math>\boxed{511}</math>. | ||
== See also == | == See also == | ||
{{AIME box|year=2004|n=I|num-b=14|after=Last Question}} | {{AIME box|year=2004|n=I|num-b=14|after=Last Question}} | ||
+ | |||
+ | [[Category:Intermediate Combinatorics Problems]] | ||
+ | [[Category:Intermediate Number Theory Problems]] |
Revision as of 13:55, 8 June 2008
Problem
For all positive integers , let
\[f(x)=\begin{cases}1 & \text{if x = 1}}\\ \frac x{10} & \text{if x is divisible by 10}\\ x+1 & \text{otherwise}\end{cases}\] (Error compiling LaTeX. Unknown error_msg)
and define a sequence as follows: and for all positive integers . Let be the smallest such that . (For example, and .) Let be the number of positive integers such that . Find the sum of the distinct prime factors of .
Solution
We backcount the number of ways. Namely, we start at , which can only be reached if , and then we perform operations that either consist of or . We represent these operations in a string format, starting with the operation that sends and so forth downwards. There are ways to pick the first operations; however, not all of them may be otherwise we return back to , contradicting our assumption that was the smallest value of . By the complement principle, we only have ways.
Since we performed the operation at least once in the first operations, it follows that , so that we no longer have to worry about reaching again. Thus the remaining operations can be picked in ways, with a total of strings.
However, we must also account for a sequence of or more s in a row, because that implies that at least one of those numbers was divisible by , yet the was never used, contradiction. We must use complement counting again by determining the number of strings of s of length such that there are s in a row. The first ten are not included since we already accounted for that scenario above, so our string of s must be preceded by a . There are no other restrictions on the remaining seven characters. Letting to denote either of the functions, and to indicate that the character appears times in a row, then our bad strings can take the forms:
&\underbrace{BA^{[10]}}\square \square \square \square \square \square \square \square \\ &\square \underbrace{BA^{[10]}}\square \square \square \square \square \square \square \\ &\qquad \quad \cdots \quad \cdots \\ &\square \square \square \square \square \square \square \underbrace{BA^{[10]}} \square \\ &\square \square \square \square \square \square \square \square \underbrace{BA^{[10]}} \\
\end{align*}$ (Error compiling LaTeX. Unknown error_msg)There are ways to select the operations for the s, and places to place our block. Thus, our answer is , and the answer is .
See also
2004 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last Question | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |