2023 AMC 10A Problems/Problem 5

Revision as of 21:04, 9 November 2023 by Not slay (talk | contribs) (Solution 1)

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 factorization of this gives us $2^{15}\cdot3^{5}\cdot5^{15}$. Pairing $2^{15}$ and $5^{15}$ together gives us a number with $15$ zeros, or 15 digits. $3^5=243$ and this adds an extra 3 digits. $15+3=\boxed{\textbf{(E) 18}}$

~zhenghua ~not_slay

See Also

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
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png