Difference between revisions of "2023 AMC 10A Problems/Problem 5"
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| + | ==Problem== | ||
| + | How many digits are in the base-ten representation of <math>8^5 \cdot 5^{10} \cdot 15^5</math>? | ||
| + | <cmath>\textbf{(A)}~14\qquad\textbf{(B)}~15\qquad\textbf{(C)}~16\qquad\textbf{(D)}~17\qquad\textbf{(E)}~18\qquad</cmath> | ||
| + | |||
| + | ==Solution 1== | ||
| + | Prime factorizing this gives us <math>2^{15}\cdot3^{5}\cdot5^{15}</math> Pairing <math>2^{15}</math> and <math>5^{15}</math> gives us a number with <math>15</math> zeros giving us 15 digits. <math>3^5=243</math> and this adds an extra 3 digits. <math>15+3=\text{\boxed{(E)18}}</math> | ||
Revision as of 16:14, 9 November 2023
Problem
How many digits are in the base-ten representation of 
?
![\[\textbf{(A)}~14\qquad\textbf{(B)}~15\qquad\textbf{(C)}~16\qquad\textbf{(D)}~17\qquad\textbf{(E)}~18\qquad\]](http://latex.artofproblemsolving.com/3/a/f/3afbcba00b7455be27438ad5aaf1f3f3fc9c6ed6.png)
Solution 1
Prime factorizing this gives us 
Pairing 
and 
gives us a number with 
zeros giving us 15 digits. 
and this adds an extra 3 digits. 
