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!</math>, we have <math>98!( | + | 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 multipls 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>. |
==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 = 26</math>. | |
| − | - | + | == 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== | ||
Latest revision as of 01:26, 22 March 2023
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 multipls of 
. Now, 
has 
factors of , so there are a total of 
factors of .
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 
.
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. 
