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| − | ==Problem==
| + | wertesryrtutyrudtu |
| − | 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> ?
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| − | <math>\textbf{(A) }23\qquad\textbf{(B) }24\qquad\textbf{(C) }25\qquad\textbf{(D) }26\qquad\textbf{(E) }27</math>
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| | ==Solution 1== | | ==Solution 1== |
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| | ~CHECKMATE2021 | | ~CHECKMATE2021 |
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| − | ==Solution 3==
| + | Note: Can you say what formula this uses? most AMC 8 test takers won't know it. Also, can someone unvandalize this page? |
| − | We can first factor a <math>98!</math> out of the <math>98! + 99! + 100!</math> to get <math>98! ( 1 + 99 + 100*99 ),</math> Simplify to get <math>98! (10,000)</math>.
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| − | Let's first find how many factors of <math>5 10,000</math> has. <math>10,000</math> is <math>(2*5)^4</math> because <math>10,000</math> is <math>(10)^4</math>. After we remove the brackets, we get <math>2^4</math>, and <math>5^4</math>. We only care about the latter (second one), because the problem only ask's for the power of <math>5</math>. We get <math>4</math>
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| − | Next, we can look at the multiples of 5 in <math>98!</math>. <math>98/5 = 19</math> so there is 19 multiples of 5. We get <math>19</math>
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| − | But we cannot forget the multiples of <math>5</math> with <math>2</math> fives in it. Multiples of <math>25</math>. How many multiples of <math>25</math> are between <math>1</math> and <math>98</math>? <math>3</math>. <math>25,50,75,</math> and that's it. We get <math>3</math>
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| − | Finally, we add all of the numbers (powers of <math>5</math>) up. That is <math>4 + 19 + 3</math>, which is just <math>26</math>
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| − | So the answer is <math>26</math>. Which is answer choice D <math>\boxed{\textbf{(D)}\ 26}</math>.
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| − | ~CHECKMATE2021
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| | ==Video Solution (CREATIVE THINKING + ANALYSIS!!!)== | | ==Video Solution (CREATIVE THINKING + ANALYSIS!!!)== |
| − | https://youtu.be/WKux87BEO1U | + | https:/90ijn bidxrfgv |
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| − | ~Education, the Study of Everything
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| − | == Video Solution by OmegaLearn==
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| − | https://youtu.be/HISL2-N5NVg?t=817
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| − | ~ pi_is_3.14
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| − | == Video Solution ==
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| − | https://youtu.be/alj9Y8jGNz8
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| − | https://youtu.be/meEuDzrM5Ac
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| − | ~savannahsolver
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| − | ==See Also==
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| − | {{AMC8 box|year=2017|num-b=18|num-a=20}}
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| − | {{MAA Notice}}
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, we have 
, which is 
. Next, 
has 
factors of 
. The 
is because of all the multiples of 
is because of all the multiples of 
. Now, 
has 
factors of 
factors of 
