2019 AMC 10A Problems/Problem 2: Difference between revisions
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== Problem == | == Problem == | ||
What is the hundreds digit of < | What is the hundreds digit of <imath>(20!-15!)?</imath> | ||
< | <imath>\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }4\qquad\textbf{(E) }5</imath> | ||
== Solution == | ==Solution 1== | ||
Because we know that <imath>5^3</imath> is a factor of <imath>15!</imath> and <imath>20!</imath>, the last three digits of both numbers is a <imath>0</imath>, this means that the difference of the hundreds digits is also <imath>\boxed{\textbf{(A) }0}</imath>. | |||
==Solution 2== | |||
We can clearly see that <imath>20! \equiv 15! \equiv 0 \pmod{1000}</imath>, so <imath>20! - 15! \equiv 0 \pmod{100}</imath> meaning that the last two digits are equal to <imath>00</imath> and the hundreds digit is <imath>\boxed{\textbf{(A)}\ 0}</imath>. | |||
20 and 15 are both | --abhinavg0627 | ||
==Solution 3 (Brute Force)== | |||
<imath>20!= 2432902008176640000</imath> | |||
and <imath>15!= 1307674368000</imath> | |||
Then, we see that the hundreds digit is <imath>0-0=\boxed{\textbf{(A)}\ 0}</imath>. | |||
~dragoon | |||
Please do not do this and only use this solution as a last resort. | |||
Note for people not used to comp. math: This is a completely reasonable way to solve this problem. | |||
==Solution 4 (Solution 1 but simpler)== | |||
The prime factorization of <imath>15!</imath> (it is easier than it seems) is <imath>2^(11) * 3^6 * 5^3 * 7^2 * 11 * 13</imath> | |||
Notice that it includes <imath>2^3 * 5^3</imath> which is 1000 | |||
Therefore 15! and 20! are both multiples of 1000, the hundreds place is <imath>\boxed{\textbf{(A)}\ 0}</imath>. | |||
==Video Solution by Education, the Study of Everything== | |||
https://youtu.be/J4Bqztwjyxw | |||
~Education, The Study of Everything | |||
==Video Solution by WhyMath== | |||
https://youtu.be/V1fY0oLSHvo | |||
~savannahsolver | |||
==Video Solution by OmegaLearn== | |||
https://youtu.be/zfChnbMGLVQ?t=3899 | |||
~pi_is_3.14 | |||
==See Also== | ==See Also== | ||
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{{AMC10 box|year=2019|ab=A|num-b=1|num-a=3}} | {{AMC10 box|year=2019|ab=A|num-b=1|num-a=3}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
[[Category: Introductory Number Theory Problems]] | |||
Latest revision as of 09:58, 10 November 2025
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)
and 
Then, we see that the hundreds digit is 
.
~dragoon
Please do not do this and only use this solution as a last resort.
Note for people not used to comp. math: This is a completely reasonable way to solve this problem.
Solution 4 (Solution 1 but simpler)
The prime factorization of (it is easier than it seems) is 
Notice that it includes 
which is 1000
Therefore 15! and 20! are both multiples of 1000, the hundreds place is .
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 | ||
These problems are copyrighted © by the Mathematical Association of America. 
