Difference between revisions of "2023 AMC 12A Problems/Problem 4"
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==Solution 1== | ==Solution 1== | ||
| − | Prime factorizing this gives us <math>2^{15}\cdot3^{5}\cdot5^{15}</math> | + | Prime factorizing this gives us <math>2^{15}\cdot3^{5}\cdot5^{15}=10^{15}\cdot3^5=243\cdot10^{15}</math>. <math>10^{15}</math> gives us <math>15</math> digits, and <math>243</math> gives us <math>3</math> digits. <math>15+3=\text{\boxed{(E) 18}}</math> |
~zhenghua | ~zhenghua | ||
==See Also== | ==See Also== | ||
| + | {{AMC10 box|year=2023|ab=A|num-b=4|num-a=6}} | ||
{{AMC12 box|year=2023|ab=A|num-b=3|num-a=5}} | {{AMC12 box|year=2023|ab=A|num-b=3|num-a=5}} | ||
| − | |||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 22:21, 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 
. 
gives us 
digits, and 
gives us 
digits. 
~zhenghua
See Also
| 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. 
