Difference between revisions of "2023 AMC 12A Problems/Problem 4"

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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 by Math-X (First understand the problem!!!)==
 +
https://youtu.be/GP-DYudh5qU?si=1RDs-j8Cedw02bID&t=983
 +
  
 
==Video Solution (easy to digest) by Power Solve==
 
==Video Solution (easy to digest) by Power Solve==
 
https://youtu.be/Od1Spf3TDBs
 
https://youtu.be/Od1Spf3TDBs
 
==Video Solution by Math-X (First understand the problem!!!)==
 
https://youtu.be/GP-DYudh5qU?si=1RDs-j8Cedw02bID&t=983
 
  
 
== Video Solution by CosineMethod [🔥Fast and Easy🔥]==
 
== Video Solution by CosineMethod [🔥Fast and Easy🔥]==

Revision as of 06:48, 6 September 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 by Math-X (First understand the problem!!!)

https://youtu.be/GP-DYudh5qU?si=1RDs-j8Cedw02bID&t=983


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

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 (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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