Difference between revisions of "2017 AMC 8 Problems/Problem 19"
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− | ==Problem | + | ==Problem== |
For any positive integer <math>M</math>, the notation <math>M!</math> denotes the product of the integers <math>1</math> through <math>M</math>. What is the largest integer <math>n</math> for which <math>5^n</math> is a factor of the sum <math>98!+99!+100!</math> ? | For any positive integer <math>M</math>, the notation <math>M!</math> denotes the product of the integers <math>1</math> through <math>M</math>. What is the largest integer <math>n</math> for which <math>5^n</math> is a factor of the sum <math>98!+99!+100!</math> ? | ||
<math>\textbf{(A) }23\qquad\textbf{(B) }24\qquad\textbf{(C) }25\qquad\textbf{(D) }26\qquad\textbf{(E) }27</math> | <math>\textbf{(A) }23\qquad\textbf{(B) }24\qquad\textbf{(C) }25\qquad\textbf{(D) }26\qquad\textbf{(E) }27</math> | ||
− | ==Solution== | + | ==Solution 1== |
− | Factoring | + | Factoring out <math>98!+99!+100!</math>, we have <math>98! (1+99+99*100)</math>, which is <math>98! (10000)</math>. Next, <math>98!</math> has <math>\left\lfloor\frac{98}{5}\right\rfloor + \left\lfloor\frac{98}{25}\right\rfloor = 19 + 3 = 22</math> factors of <math>5</math>. The <math>19</math> is because of all the multiples of <math>5</math>.The <math>3</math> is because of all the multiples of <math>25</math>. Now, <math>10,000</math> has <math>4</math> factors of <math>5</math>, so there are a total of <math>22 + 4 = \boxed{\textbf{(D)}\ 26}</math> factors of <math>5</math>. |
+ | |||
+ | ~CHECKMATE2021 | ||
+ | |||
+ | ==Video Solution (CREATIVE THINKING + ANALYSIS!!!)== | ||
+ | https://youtu.be/WKux87BEO1U | ||
+ | |||
+ | ~Education, the Study of Everything | ||
==See Also== | ==See Also== |
Revision as of 22:34, 24 May 2024
Problem
For any positive integer , the notation denotes the product of the integers through . What is the largest integer for which is a factor of the sum ?
Solution 1
Factoring out , we have , which is . Next, has factors of . The is because of all the multiples of .The is because of all the multiples of . Now, has factors of , so there are a total of factors of .
~CHECKMATE2021
Video Solution (CREATIVE THINKING + ANALYSIS!!!)
~Education, the Study of Everything
See Also
2017 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
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All AJHSME/AMC 8 Problems and Solutions |
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