Difference between revisions of "2023 AMC 10A Problems/Problem 5"
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==Solution 1== | ==Solution 1== | ||
| − | Prime factorization of 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, | + | Prime factorization of this gives us <math>2^{15}\cdot3^{5}\cdot5^{15}</math>. Pairing <math>2^{15}</math> and <math>5^{15}</math> together gives us a number with <math>15</math> zeros, or 15 digits. <math>3^5=243</math> and this adds an extra 3 digits. <math>15+3=\boxed{\textbf{(E) 18}}</math> |
~zhenghua | ~zhenghua | ||
Revision as of 20:52, 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 factorization of this gives us 
. Pairing 
and 
together gives us a number with 
zeros, or 15 digits. 
and this adds an extra 3 digits. 
~zhenghua
See Also
| 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 | |
| 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 | ||
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