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> has <math>16</math> digits and <math>243</math> = <math>2.43*10^{2}</math> gives us <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 | <math>2.43*10^{17}</math> has <math>18</math> digits | ||
| Line 15: | Line 15: | ||
~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)== |
| − | + | 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== | ==Video Solution (easy to digest) by Power Solve== | ||
Latest 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
- 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. 
