Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m
(→‎Solution 1)
Line 4: Line 4:
 
\frac{n}{3}</math>. If <math>n</math> is not a multiple of <math>3</math>, then you replace <math>n</math> by <math>n+10</math>. Then continue this process. For example, beginning with <math>n=4</math>, this procedure gives <math>4 \rightarrow 14 \rightarrow 24 \rightarrow 8 \rightarrow 18 \rightarrow 6 \rightarrow 2 \rightarrow 12 \rightarrow \cdots</math>. Suppose you start with <math>n=100</math>. What value results if you perform this operation exactly <math>100</math> times?
 
\frac{n}{3}</math>. If <math>n</math> is not a multiple of <math>3</math>, then you replace <math>n</math> by <math>n+10</math>. Then continue this process. For example, beginning with <math>n=4</math>, this procedure gives <math>4 \rightarrow 14 \rightarrow 24 \rightarrow 8 \rightarrow 18 \rightarrow 6 \rightarrow 2 \rightarrow 12 \rightarrow \cdots</math>. Suppose you start with <math>n=100</math>. What value results if you perform this operation exactly <math>100</math> times?
  
−
== Solution 1 ==
+
== Solution 1 (fast) ==
 
Let <math>s</math> be the number of times the operation is performed. Notice the sequence goes <math>100 \rightarrow 110 \rightarrow 120 \rightarrow 40 \rightarrow 50 \rightarrow 60 \rightarrow 20 \rightarrow 30 \rightarrow 10 \rightarrow 20 \rightarrow \cdots</math>. Thus, for <math>s \equiv 1 \pmod{3}</math>, the value is <math>30</math>. Since <math>100 \equiv 1 \pmod{3}</math>, the answer is <math>\boxed{\textbf{(C)} 30}</math>
 
Let <math>s</math> be the number of times the operation is performed. Notice the sequence goes <math>100 \rightarrow 110 \rightarrow 120 \rightarrow 40 \rightarrow 50 \rightarrow 60 \rightarrow 20 \rightarrow 30 \rightarrow 10 \rightarrow 20 \rightarrow \cdots</math>. Thus, for <math>s \equiv 1 \pmod{3}</math>, the value is <math>30</math>. Since <math>100 \equiv 1 \pmod{3}</math>, the answer is <math>\boxed{\textbf{(C)} 30}</math>
  
 
~andliu766
 
~andliu766
 +
 +
== Solution 2 (more explanatory) ==
 +
Looking at the first few values of our operation, we get <math>100 \rightarrow 110 \rightarrow 120 \rightarrow 40 \rightarrow 50 \rightarrow 60 \rightarrow 20 \rightarrow 30 \rightarrow 10 \rightarrow 20</math>. We can see that <math>30</math> will go to <math>10</math>, then to <math>20</math>, then back to <math>30</math>, and the loop resets. After 7 operations, we reach <math>30</math>. We still have 93 operations left, so because the loop will run exactly <math>31</math> times <math>(93/3)</math>, we will reach at <math>30</math> again. So,, the answer is <math>\boxed{\textbf{(C)} 30}</math>
 +
 +
edit for clarity pls
  
 
==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 23:22, 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

See also

2024 AMC 10A (Problems • Answer Key • Resources)
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

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2024_AMC_10A_Problems/Problem_10&oldid=232427"