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

Revision as of 09:42, 9 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 $4 \to 14 \to 24 \to 8 \to 18 \to 6 \to 2 \to 12 \to \cdots$ . 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 ⚡️⚡️⚡️)

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 $30$ 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 grammar pls

~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 $100-6$ moves to go, so to find what number we land on after $94$ more steps, we can do $94 \pmod {3} \equiv (9 + 4) \pmod {3} = 1$ , meaning we go from $20 \to \boxed{\textbf{(C) } 30}$ .

~yuvag

~alot of credit to Moonwatcher22

Video Solution 1 by Power Solve

https://youtu.be/wamtu7xm0eU

@@ Line 19: / Line 19: @@
 == 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>100-6</math> moves to go, so to find what number we land on after <math>94</math> more steps, we can do  <math>94 \pmod {3} \equiv (9 + 4) \pmod {3} = 1</math>, meaning we go from <math>20 \to </math>\boxed{\textbf{(C) } 30}$.
+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>100-6</math> moves to go, so to find what number we land on after <math>94</math> more steps, we can do  <math>94 \pmod {3} \equiv (9 + 4) \pmod {3} = 1</math>, meaning we go from <math>20 \to \boxed{\textbf{(C) } 30}</math>.
 ~yuvag

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