Difference between revisions of "2023 AMC 12A Problems/Problem 4"
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==Problem== | ==Problem== | ||
− | How many digits are in the base-ten representation of <math>8^5 \cdot 5^{10} \cdot 15^ | + | How many digits are in the base-ten representation of <math>8^5 \cdot 5^{10} \cdot 15^2</math>? |
− | <math>\textbf{(A)}~14\qquad\textbf{(B)}~15\qquad\textbf{(C)}~16\qquad\ | + | <math>\textbf{(A)}~14\qquad\textbf{(B)}~15\qquad\textbf{(C)}~16\qquad\texbf{(D)}~17\qquad\textbf{(E)}~18</math> |
==Solution 1== | ==Solution 1== | ||
Prime factorizing this gives us <math>2^{15}\cdot3^{5}\cdot5^{15}=10^{15}\cdot3^5=243\cdot10^{15}</math>. | Prime factorizing this gives us <math>2^{15}\cdot3^{5}\cdot5^{15}=10^{15}\cdot3^5=243\cdot10^{15}</math>. | ||
− | <math>10^{15}</math> | + | <math>10^{15}</math> has <math>16</math> digits and <math>243</math> = <math>2.43*10^{2}</math> gives us <math>3</math> more digits. <math>16+2=\text{\boxed{\textbf{(E) }18}}</math> |
+ | |||
+ | <math>2.43*10^{17}</math> has <math>18</math> digits | ||
~zhenghua | ~zhenghua | ||
− | ==Solution 2== | + | ==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 <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 | ||
− | == | + | ==Solution 3 (Similar to Solution 1)== |
− | https://youtu.be/ | + | All the exponents have a common factor of <math>5</math> which we can factor out. This leaves us with <math>(8 \cdot 5^2 \cdot 15)^5 = (3000)^5 = (3 \cdot 1000)^5</math>. We can then distribute the power leaving us with <math>3^5 \cdot 10^{3 \cdot 5} = 243 \cdot 10^{15}</math>. This would be <math>243</math> followed by <math>15</math> zeros resulting in our answer being <math>15+3=\text{\boxed{\textbf{(E)}18}}</math> |
+ | |||
+ | ~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== | ||
+ | https://youtu.be/Od1Spf3TDBs | ||
+ | |||
+ | == Video Solution by CosineMethod [🔥Fast and Easy🔥]== | ||
+ | |||
+ | https://www.youtube.com/watch?v=laHiorWO1zo | ||
==Video Solution== | ==Video Solution== | ||
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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
− | ==Video Solution ( | + | ==Video Solution (⚡ Under 2 min ⚡)== |
https://youtu.be/Xy8vyymlPBg | https://youtu.be/Xy8vyymlPBg | ||
Revision as of 18:50, 13 November 2024
- 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.