Difference between revisions of "2019 AMC 10A Problems/Problem 2"
Scrabbler94 (talk | contribs) (Solution 2 is wrong since divisibility by 100 doesn't imply the hundreds digit of their difference is 0.) |
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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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==Video Solution== | ==Video Solution== |
Revision as of 00:41, 1 July 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 .
Video Solution
~savannahsolver
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 |
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