Difference between revisions of "2019 AMC 10A Problems/Problem 2"

Latest revision as of 18:14, 4 November 2024

Problem

What is the hundreds digit of $(20!-15!)?$

$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }4\qquad\textbf{(E) }5$

Solution 1

Because we know that $5^3$ is a factor of $15!$ and $20!$ , the last three digits of both numbers is a $0$ , this means that the difference of the hundreds digits is also $\boxed{\textbf{(A) }0}$ .

Solution 2

We can clearly see that $20! \equiv 15! \equiv 0 \pmod{1000}$ , so $20! - 15! \equiv 0 \pmod{100}$ meaning that the last two digits are equal to $00$ and the hundreds digit is $\boxed{\textbf{(A)}\ 0}$ .

--abhinavg0627

Solution 3 (Brute Force)

$20!= 2432902008176640000$ $15!= 1307674368000$

Then, we see that the hundreds digit is $0-0=\boxed{\textbf{(A)}\ 0}$ .

~dragoon

Please do not do this and only use this solution as a last resort.

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

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.

@@ Line 2: / Line 2: @@
 What is the hundreds digit of <math>(20!-15!)?</math>
+<math>\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }4\qquad\textbf{(E) }5</math>
+==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 <math>0</math>, this means that the difference of the hundreds digits is also <math>\boxed{\textbf{(A) }0}</math>.
-== Solution ==
+==Solution 2==
-The last three digit of <math>n!</math> for all <math>n>=15</math> is <math>000</math>, because there are at least three 2s and three 5s in its prime factorization. Because <math>0-0=0</math>, The answer is <math>0</math>.
+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
+==Solution 3 (Brute Force)==
+<math>20!= 2432902008176640000</math>
+<math>15!= 1307674368000</math>
+Then, we see that the hundreds digit is <math>0-0=\boxed{\textbf{(A)}\ 0}</math>.
+~dragoon
+Please do not do this and only use this solution as a last resort.
+==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==
+{{AMC10 box|year=2019|ab=A|num-b=1|num-a=3}}
+{{MAA Notice}}

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