Difference between revisions of "2024 AMC 10A Problems/Problem 10"

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edit for clarity pls
 
edit for clarity pls
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~Moonwatcher22
  
 
==See also==
 
==See also==
 
{{AMC10 box|year=2024|ab=A|num-b=9|num-a=11}}
 
{{AMC10 box|year=2024|ab=A|num-b=9|num-a=11}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 23:23, 8 November 2024

Problem

Consider the following operation. Given a positive integer $n$, if $n$ is a multiple of $3$, then you replace $n$ by $\frac{n}{3}$. If $n$ is not a multiple of $3$, then you replace $n$ by $n+10$. Then continue this process. For example, beginning with $n=4$, this procedure gives $4 \rightarrow 14 \rightarrow 24 \rightarrow 8 \rightarrow 18 \rightarrow 6 \rightarrow 2 \rightarrow 12 \rightarrow \cdots$. Suppose you start with $n=100$. What value results if you perform this operation exactly $100$ times?

Solution 1 (fast)

Let $s$ be the number of times the operation is performed. Notice the sequence goes $100 \rightarrow 110 \rightarrow 120 \rightarrow 40 \rightarrow 50 \rightarrow 60 \rightarrow 20 \rightarrow 30 \rightarrow 10 \rightarrow 20 \rightarrow \cdots$. Thus, for $s \equiv 1 \pmod{3}$, the value is $30$. Since $100 \equiv 1 \pmod{3}$, the answer is $\boxed{\textbf{(C)} 30}$

~andliu766

Solution 2 (more explanatory)

Looking at the first few values of our operation, we get $100 \rightarrow 110 \rightarrow 120 \rightarrow 40 \rightarrow 50 \rightarrow 60 \rightarrow 20 \rightarrow 30 \rightarrow 10 \rightarrow 20$. We can see that $30$ will go to $10$, then to $20$, then back to $30$, and the loop resets. After 7 operations, we reach $30$. We still have 93 operations left, so because the loop will run exactly $31$ times $(93/3)$, we will reach at $30$ again. So,, the answer is $\boxed{\textbf{(C)} 30}$

edit for clarity pls

~Moonwatcher22

See also

2024 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

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