2023 AMC 12A Problems/Problem 4

Revision as of 19:07, 18 October 2025 by Ritwikrule (talk | contribs) (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))
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 $8^5 \cdot 5^{10} \cdot 15^5$?

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

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 $3$ more digits. $16+2=\text{\boxed{\textbf{(E) }18}}$

$2.43*10^{17}$ has $18$ digits

~zhenghua

Solution 3 (Similar to Solution 1)

All the exponents have a common factor of $5$ which we can factor out. This leaves us with $(8 \cdot 5^2 \cdot 15)^5 = (3000)^5 = (3 \cdot 1000)^5$. We can then distribute the power leaving us with $3^5 \cdot 10^{3 \cdot 5} = 243 \cdot 10^{15}$. This would be $243$ followed by $15$ zeros resulting in our answer being $15+3=\text{\boxed{\textbf{(E)}18}}$

~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

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

Video Solution Fast and Easy by DR.GOOGLE

https://www.youtube.com/watch?v=Q6-MiP6ZTYs

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

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