2023 AMC 12A Problems/Problem 4: Difference between revisions
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Multiplying it out, we get that <math>8^5 \cdot 5^{10} \cdot 15^5 = 243000000000000000</math>. Counting, we have the answer is <math>\text{\boxed{\textbf{(E) }18}}</math> | Multiplying it out, we get that <math>8^5 \cdot 5^{10} \cdot 15^5 = 243000000000000000</math>. Counting, we have the answer is <math>\text{\boxed{\textbf{(E) }18}}</math> | ||
~andliu766 | ~andliu766 | ||
==Video Solution (easy to digest) by Power Solve== | |||
https://youtu.be/Od1Spf3TDBs | |||
==Video Solution by Math-X (First understand the problem!!!)== | ==Video Solution by Math-X (First understand the problem!!!)== | ||
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~Education, the Study of Everything | ~Education, the Study of Everything | ||
==See Also== | ==See Also== | ||
Revision as of 19:32, 10 June 2024
- The following problem is from both the 2023 AMC 10A #5 and 2023 AMC 12A #4, so both problems redirect to this page.
Problem
How many digits are in the base-ten representation of 
?
Solution 1
Prime factorizing this gives us 
.
has 
digits and 
= 
gives us 
more digits. 
has 
digits
~zhenghua
Solution 2 (Bash)(Only if you don't know how to do the rest of the problems and have about 20 minutes left)
Multiplying it out, we get that 
. Counting, we have the answer is 
~andliu766
Video Solution (easy to digest) by Power Solve
Video Solution by Math-X (First understand the problem!!!)
https://youtu.be/cMgngeSmFCY?si=b1khjbMn1i5rApCe&t=903
Video Solution by CosineMethod [🔥Fast and Easy🔥]
https://www.youtube.com/watch?v=laHiorWO1zo
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution (⚡ Under 2 min ⚡)
~Education, the Study of Everything
See Also
| 2023 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 4 |
Followed by Problem 6 | |
| 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 | ||
| 2023 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 3 |
Followed by Problem 5 |
| 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 12 Problems and Solutions | |
These problems are copyrighted © by the Mathematical Association of America. 
