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

Latest revision as of 14:52, 24 November 2024

1 Problem
2 Solution 1 (Fast Solution)
3 Solution 2 (More Explanatory)
4 Solution 3 (very slightly different than previous)
5 Video Solution by Pi Academy
6 Video Solution 1 by Power Solve
7 Video Solution by Daily Dose of Math
8 Video Solution by SpreadTheMathLove
9 Video Solution by Just Math⚡
10 See also

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 $4 \to 14 \to 24 \to 8 \to 18 \to 6 \to 2 \to 12 \to \cdots$ . Suppose you start with $n=100$ . What value results if you perform this operation exactly $100$ times?

$\textbf{(A) }10\qquad\textbf{(B) }20\qquad\textbf{(C) }30\qquad\textbf{(D) }40\qquad\textbf{(E) }50$

Solution 1 (Fast Solution)

Let $s$ be the number of times the operation is performed. Notice the sequence goes $100 \to 110 \to 120 \to 40 \to 50 \to 60 \to 20 \to 30 \to 10 \to 20 \to \cdots$ . 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 $100 \to 110 \to 120 \to 40 \to 50 \to 60 \to 20 \to 30 \to 10 \to 20$ . We can see that $30$ goes 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 $30$ again. So, the answer is $\boxed{\textbf{(C) } 30}$ .

~Moonwatcher22

Solution 3 (very slightly different than previous)

Calculating the first few values, we get $100 \to 110 \to 120 \to 40 \to 50 \to 60 \to 20 \to 30 \to 10 \to 20$ . We can see that $20$ will go to $30$ , then to $10$ , then back to $20$ , and then the loop resets. After $6$ moves, we reach $20$ , the start of the cycle. We still have $100-5$ moves to go, so to find what number we land on after $95$ more steps, we can do $95 \pmod {3} \equiv (9 + 4) \pmod {3} = 1$ , meaning we go from $20 \to \boxed{\textbf{(C) } 30}$ .

~yuvag

~a lot of credit to Moonwatcher22

Video Solution by Pi Academy

https://youtu.be/6qYaJsgqkbs?si=K2Ebwqg-Ro8Yqoiv

Video Solution 1 by Power Solve

https://youtu.be/wamtu7xm0eU

Video Solution by Daily Dose of Math

https://youtu.be/17-o9qiprI0

~Thesmartgreekmathdude

Video Solution by SpreadTheMathLove

https://www.youtube.com/watch?v=_o5zagJVe1U

Video Solution by Just Math⚡

https://www.youtube.com/watch?v=lqZUYJPq_Jo

@@ Line 1: / Line 1: @@
+== Problem ==
+Consider the following operation. Given a positive integer <math>n</math>, if <math>n</math> is a multiple of <math>3</math>, then you replace <math>n</math> by <math>
+\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 \to 14 \to 24 \to 8 \to 18 \to 6 \to 2 \to 12 \to \cdots</math>. Suppose you start with <math>n=100</math>. What value results if you perform this operation exactly <math>100</math> times?
+<math>\textbf{(A) }10\qquad\textbf{(B) }20\qquad\textbf{(C) }30\qquad\textbf{(D) }40\qquad\textbf{(E) }50</math>
+== Solution 1 (Fast Solution) ==
+Let <math>s</math> be the number of times the operation is performed. Notice the sequence goes <math>100 \to 110 \to 120 \to 40 \to 50 \to 60 \to 20 \to 30 \to 10 \to 20 \to \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
+== Solution 2 (More Explanatory) ==
+Looking at the first few values of our operation, we get <math>100 \to 110 \to 120 \to 40 \to 50 \to 60 \to 20 \to 30 \to 10 \to 20</math>. We can see that <math>30</math> goes 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 <math>30</math> again. So, the answer is <math>\boxed{\textbf{(C) } 30}</math>.
+~Moonwatcher22
+== Solution 3 (very slightly different than previous) ==
+Calculating the first few values, we get <math>100 \to 110 \to 120 \to 40 \to 50 \to 60 \to 20 \to 30 \to 10 \to 20</math>. We can see that <math>20</math> will go to <math>30</math>, then to <math>10</math>, then back to <math>20</math>, and then the loop resets. After <math>6</math> moves, we reach <math>20</math>, the start of the cycle. We still have <math>100-5</math> moves to go, so to find what number we land on after <math>95</math> more steps, we can do  <math>95 \pmod {3} \equiv (9 + 4) \pmod {3} = 1</math>, meaning we go from <math>20 \to \boxed{\textbf{(C) } 30}</math>.
+~yuvag
+~a lot of credit to Moonwatcher22
+== Video Solution by Pi Academy ==
+https://youtu.be/6qYaJsgqkbs?si=K2Ebwqg-Ro8Yqoiv
+==Video Solution 1 by Power Solve ==
+https://youtu.be/wamtu7xm0eU
+== Video Solution by Daily Dose of Math ==
+https://youtu.be/17-o9qiprI0
+~Thesmartgreekmathdude
+==Video Solution by SpreadTheMathLove==
+https://www.youtube.com/watch?v=_o5zagJVe1U
+==Video Solution by Just Math⚡==
+https://www.youtube.com/watch?v=lqZUYJPq_Jo
+==See also==
+{{AMC10 box|year=2024|ab=A|num-b=9|num-a=11}}
+{{MAA Notice}}

2024 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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