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> ? | ||
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==Solution 1== | ==Solution 1== | ||
− | 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>. 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>. | + | 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>. |
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+ | ~CHECKMATE2021 | ||
==Solution 2== | ==Solution 2== | ||
− | + | Also, keep in mind that the number of <math>5</math>’s in <math>98! (10,000)</math> is the same as the number of trailing zeros. The number of zeros is <math>98!</math>, which means we need pairs of <math>5</math>’s and <math>2</math>’s; we know there will be many more <math>2</math>’s, so we seek to find the number of <math>5</math>’s in <math>98!</math>, which the solution tells us. And, that is <math>22</math> factors of <math>5</math>. <math>10,000</math> has <math>4</math> trailing zeros, so it has <math>4</math> factors of <math>5</math> and <math>22 + 4 = \boxed{26}</math>. | |
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+ | ~CHECKMATE2021 | ||
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==Solution 3== | ==Solution 3== | ||
− | We can | + | 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 | ||
+ | |||
+ | ==Video Solution (CREATIVE THINKING + ANALYSIS!!!)== | ||
+ | https://youtu.be/WKux87BEO1U | ||
+ | |||
+ | ~Education, the Study of Everything | ||
+ | |||
+ | == Video Solution by OmegaLearn== | ||
+ | https://youtu.be/HISL2-N5NVg?t=817 | ||
− | + | ~ pi_is_3.14 | |
− | |||
− | + | == Video Solution == | |
+ | https://youtu.be/alj9Y8jGNz8 | ||
+ | https://youtu.be/meEuDzrM5Ac | ||
− | + | ~savannahsolver | |
− | |||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2017|num-b=18|num-a=20}} | {{AMC8 box|year=2017|num-b=18|num-a=20}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 18:18, 21 January 2024
Contents
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
Solution 2
Also, keep in mind that the number of ’s in is the same as the number of trailing zeros. The number of zeros is , which means we need pairs of ’s and ’s; we know there will be many more ’s, so we seek to find the number of ’s in , which the solution tells us. And, that is factors of . has trailing zeros, so it has factors of and .
~CHECKMATE2021
Solution 3
We can first factor a out of the to get Simplify to get .
Let's first find how many factors of has. is because is . After we remove the brackets, we get , and . We only care about the latter (second one), because the problem only ask's for the power of . We get
Next, we can look at the multiples of 5 in . so there is 19 multiples of 5. We get
But we cannot forget the multiples of with fives in it. Multiples of . How many multiples of are between and ? . and that's it. We get
Finally, we add all of the numbers (powers of ) up. That is , which is just
So the answer is . Which is answer choice D .
~CHECKMATE2021
Video Solution (CREATIVE THINKING + ANALYSIS!!!)
~Education, the Study of Everything
Video Solution by OmegaLearn
https://youtu.be/HISL2-N5NVg?t=817
~ pi_is_3.14
Video Solution
~savannahsolver
See Also
2017 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
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 AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.