Difference between revisions of "2024 AMC 10A Problems/Problem 10"
(→Solution 2 (More Explanatory)) |
|||
Line 17: | Line 17: | ||
~Moonwatcher22 | ~Moonwatcher22 | ||
+ | |||
+ | == Solution 3 (Slightly different than previous) == | ||
+ | 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> will go to <math>10</math>, then to <math>20</math>, then back to <math>30</math>, and the loop resets. After 6 operations, we reach <math>20</math>, the start of the cycle. We still have <math>100-6</math> operations left, so to find what we land on after <math>94 more steps, we can do </math>94 (\mod {3}) = (9 + 4) (\mod {3}) = 1<math>, meaning we go from </math>20 \to \boxed{30}<math>. So, the answer is </math>\boxed{\textbf{(C) } 30}$. | ||
+ | |||
+ | ~yuvag | ||
== Video Solution 1 by Power Solve == | == Video Solution 1 by Power Solve == |
Revision as of 09:28, 9 November 2024
Contents
Problem
Consider the following operation. Given a positive integer , if is a multiple of , then you replace by . If is not a multiple of , then you replace by . Then continue this process. For example, beginning with , this procedure gives . Suppose you start with . What value results if you perform this operation exactly times?
Solution 1 (fast ⚡️⚡️⚡️)
Let be the number of times the operation is performed. Notice the sequence goes . Thus, for , the value is . Since , the answer is .
~andliu766
Solution 2 (More Explanatory)
Looking at the first few values of our operation, we get . We can see that will go to , then to , then back to , and the loop resets. After 7 operations, we reach . We still have 93 operations left, so because the loop will run exactly times , we will reach at again. So, the answer is .
edit for grammar pls
~Moonwatcher22
Solution 3 (Slightly different than previous)
Looking at the first few values of our operation, we get . We can see that will go to , then to , then back to , and the loop resets. After 6 operations, we reach , the start of the cycle. We still have operations left, so to find what we land on after 94 (\mod {3}) = (9 + 4) (\mod {3}) = 120 \to \boxed{30}\boxed{\textbf{(C) } 30}$.
~yuvag
Video Solution 1 by Power Solve
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.