Difference between revisions of "2024 AMC 10A Problems/Problem 10"
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+ | == Video Solution by Daily Dose of Math == | ||
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+ | https://youtu.be/17-o9qiprI0 | ||
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+ | ~Thesmartgreekmathdude | ||
==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 10:23, 10 November 2024
Contents
[hide]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 Solution)
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 goes 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 again. So, the answer is .
~Moonwatcher22
Solution 3 (very slightly different than previous)
Calculating the first few values, we get . We can see that will go to , then to , then back to , and then the loop resets. After moves, we reach , the start of the cycle. We still have moves to go, so to find what number we land on after more steps, we can do , meaning we go from .
~yuvag
~alot of credit to Moonwatcher22
Video Solution by Pi Academy
https://youtu.be/6qYaJsgqkbs?si=K2Ebwqg-Ro8Yqoiv
Video Solution 1 by Power Solve
Video Solution by Daily Dose of Math
~Thesmartgreekmathdude
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.