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

Revision as of 14:41, 11 November 2023

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 $2^{15}\cdot3^{5}\cdot5^{15}=10^{15}\cdot3^5=243\cdot10^{15}$ . $10^{15}$ gives us $15$ digits, and $243$ gives us $3$ digits. $15+3=\text{\boxed{\textbf{(E) }18}}$

~zhenghua

Video Solution by Math-X (First understand the problem!!!)

https://youtu.be/cMgngeSmFCY?si=b1khjbMn1i5rApCe&t=903

Video Solution

https://youtu.be/MFPSxqtguQo

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

Video Solution (Quick and Easy!)

https://youtu.be/Xy8vyymlPBg

~Education, the Study of Everything

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.

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