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

m (Problem)
(Problem)
(2 intermediate revisions by 2 users not shown)
Line 4: Line 4:
 
How many digits are in the base-ten representation of <math>8^5 \cdot 5^{10} \cdot 15^2</math>?
 
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\texf{(D)}~17\qquad\textbf{(E)}~18\qquad</math>
+
<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==
Line 27: Line 27:
 
https://youtu.be/h2Pf2hvF1wE?si=mJjq7vPLptdSe0AJ&t=872
 
https://youtu.be/h2Pf2hvF1wE?si=mJjq7vPLptdSe0AJ&t=872
 
~little-fermat
 
~little-fermat
 
==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==

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.

Problem

How many digits are in the base-ten representation of $8^5 \cdot 5^{10} \cdot 15^2$?

$\textbf{(A)}~14\qquad\textbf{(B)}~15\qquad\textbf{(C)}~16\qquad\texbf{(D)}~17\qquad\textbf{(E)}~18$ (Error compiling LaTeX. Unknown error_msg)

Solution 1

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

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

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

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