2023 AMC 12A Problems/Problem 4
- The following problem is from both the 2023 AMC 10A #5 and 2023 AMC 12A #4, so both problems redirect to this page.
Contents
[hide]- 1 Problem
- 2 Solution 1
- 3 Solution 2 (Only if you don't know how to do the rest of the problems and have about 20 minutes left, IT IS recommended)
- 4 Solution 3 (Similar to Solution 1)
- 5 Video Solution by Little Fermat
- 6 Video Solution (easy to digest) by Power Solve
- 7 Video Solution by CosineMethod [🔥Fast and Easy🔥]
- 8 Video Solution
- 9 Video Solution (⚡ Under 2 min ⚡)
- 10 See Also
Problem
How many digits are in the base-ten representation of ?
$\textbf{(A)}~14\qquad\textbf{(B)}~15\qquad\textbf{(C)}~16\qquad\texbf{(D)}~17\qquad\textbf{(E)}~18$ (Error compiling LaTeX. Unknown error_msg)
Solution 1
Prime factorizing this gives us .
has digits and = gives us more digits.
has digits
~zhenghua
Solution 2 (Only if you don't know how to do the rest of the problems and have about 20 minutes left, IT IS recommended)
Multiplying it out, we get that . Counting, we have the answer is ~andliu766
Solution 3 (Similar to Solution 1)
All the exponents have a common factor of which we can factor out. This leaves us with . We can then distribute the power leaving us with . This would be followed by zeros resulting in our answer being
~leon_0iler
Video Solution by Little Fermat
https://youtu.be/h2Pf2hvF1wE?si=mJjq7vPLptdSe0AJ&t=872 ~little-fermat
Video Solution (easy to digest) by Power Solve
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.