Difference between revisions of "2019 AMC 10A Problems/Problem 2"
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<math>\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }4\qquad\textbf{(E) }5</math> | <math>\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }4\qquad\textbf{(E) }5</math> | ||
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+ | ==Solution 1== | ||
+ | Because we know that <math>5^3</math> is a factor of <math>15!</math> and <math>20!</math>, the last three digits of both numbers is a 0, this means that the difference of the hundreds digits is also <math>\boxed{\textbf{(A) }0}</math>. | ||
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+ | ==Solution 2== | ||
+ | |||
+ | We can clearly see that <math>20! \equiv 15! \equiv 0 \pmod{1000}</math>, so <math>20! - 15! \equiv 0 \pmod{100}</math> meaning that the last two digits are equal to <math>00</math> and the hundreds digit is <math>0</math>, or <math>\boxed{\textbf{(A)}\ 0}</math>. | ||
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+ | --abhinavg0627 | ||
+ | |||
+ | ==Solution 3 (Brute Force)== | ||
+ | |||
+ | <math>20!= 2432902008176640000</math> | ||
+ | <math>15!= 1307674368000</math> | ||
+ | |||
+ | Then, we see that the hundred digit is <math>0-0=\boxed{\textbf{(A)}\ 0}</math>. | ||
+ | |||
+ | ~dragoon | ||
==Video Solution 1== | ==Video Solution 1== | ||
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==See Also== | ==See Also== |
Revision as of 11:57, 16 July 2024
Contents
[hide]Problem
What is the hundreds digit of
Solution 1
Because we know that is a factor of and , the last three digits of both numbers is a 0, this means that the difference of the hundreds digits is also .
Solution 2
We can clearly see that , so meaning that the last two digits are equal to and the hundreds digit is , or .
--abhinavg0627
Solution 3 (Brute Force)
Then, we see that the hundred digit is .
~dragoon
Video Solution 1
Education, The Study of Everything
Video Solution 2
~savannahsolver
Video Solution 3 by OmegaLearn
https://youtu.be/zfChnbMGLVQ?t=3899
~ pi_is_3.14
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.