Difference between revisions of "2019 AMC 10A Problems/Problem 2"
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The last three digits of <math>n!</math> for all <math>n\geq15</math> are <math>000</math>, because there are at least three <math>2</math>s and three <math>5</math>s in its prime factorization. Because <math>0-0=0</math>, the answer is <math>\boxed{\textbf{(A) }0}</math>. | The last three digits of <math>n!</math> for all <math>n\geq15</math> are <math>000</math>, because there are at least three <math>2</math>s and three <math>5</math>s in its prime factorization. Because <math>0-0=0</math>, the answer is <math>\boxed{\textbf{(A) }0}</math>. | ||
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| + | == Solution 2 == | ||
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| + | 20 and 15 are both greater than 10, they are divisible by 100, and therefore, the hundreds digit is <math>\boxed{\textbf{(A) }0}</math>. ~ Peppapig_ | ||
==See Also== | ==See Also== | ||
Revision as of 20:55, 1 April 2020
Contents
Problem
What is the hundreds digit of 
Solution
The last three digits of 
for all 
are 
, because there are at least three 
s and three 
s in its prime factorization. Because 
, the answer is 
.
Solution 2
20 and 15 are both greater than 10, they are divisible by 100, and therefore, the hundreds digit is . ~ Peppapig_
See Also
| 2019 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 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. 
