Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2023 AMC 12A Problems/Problem 4"
(13 intermediate revisions by 8 users not shown)
Line 7: Line 7:
  
 
==Solution 1==
 
==Solution 1==
−
Prime factorizing this gives us <math>2^{15}\cdot3^{5}\cdot5^{15}=10^{15}\cdot3^5=243\cdot10^{15}</math>. <math>10^{15}</math> gives us <math>15</math> digits, and <math>243</math> gives us <math>3</math> digits. <math>15+3=\text{\boxed{\textbf{(E) }18}}</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>2</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 (Bash)(Only if you don't know how to do the rest of the problems and have about 20 minutes left)==
−
Using a calculator, 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
  
 +
==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!!!)==
 
https://youtu.be/cMgngeSmFCY?si=b1khjbMn1i5rApCe&t=903
 
https://youtu.be/cMgngeSmFCY?si=b1khjbMn1i5rApCe&t=903
 +
 +
== Video Solution by CosineMethod [🔥Fast and Easy🔥]==
 +
 +
https://www.youtube.com/watch?v=laHiorWO1zo
  
 
==Video Solution==
 
==Video Solution==
Line 24: Line 35:
 
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
  
−
==Video Solution (Quick and Easy!)==
+
==Video Solution (⚡ Under 2 min ⚡)==
 
https://youtu.be/Xy8vyymlPBg
 
https://youtu.be/Xy8vyymlPBg
  

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 $8^5 \cdot 5^{10} \cdot 15^5$?

$\textbf{(A)}~14\qquad\textbf{(B)}~15\qquad\textbf{(C)}~16\qquad\textbf{(D)}~17\qquad\textbf{(E)}~18\qquad$

Solution 1

Prime factorizing this gives us $2^{15}\cdot3^{5}\cdot5^{15}=10^{15}\cdot3^5=243\cdot10^{15}$.

$10^{15}$ has $16$ digits and $243$ = $2.43*10^{2}$ gives us $2$ more digits. $16+2=\text{\boxed{\textbf{(E) }18}}$

$2.43*10^{17}$ has $18$ 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 $8^5 \cdot 5^{10} \cdot 15^5 = 243000000000000000$. Counting, we have the answer is $\text{\boxed{\textbf{(E) }18}}$ ~andliu766

Video Solution (easy to digest) by Power Solve

https://youtu.be/Od1Spf3TDBs

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

https://youtu.be/MFPSxqtguQo

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Video Solution (⚡ Under 2 min ⚡)

https://youtu.be/Xy8vyymlPBg

~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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2023_AMC_12A_Problems/Problem_4&oldid=220335"