Difference between revisions of "2019 AMC 10A Problems/Problem 2"
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==Solution 2== | ==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 | + | 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>\boxed{\textbf{(A)}\ 0}</math>. |
--abhinavg0627 | --abhinavg0627 | ||
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<math>15!= 1307674368000</math> | <math>15!= 1307674368000</math> | ||
| − | Then, we see that the | + | Then, we see that the hundreds digit is <math>0-0=\boxed{\textbf{(A)}\ 0}</math>. |
~dragoon | ~dragoon | ||
Latest revision as of 12:59, 16 July 2024
Contents
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 
, 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 
.
--abhinavg0627
Solution 3 (Brute Force)
Then, we see that the hundreds digit is 
.
~dragoon
Video Solution by Education, the Study of Everything
~Education, The Study of Everything
Video Solution by WhyMath
~savannahsolver
Video Solution 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. 
