Difference between revisions of "2017 AMC 8 Problems/Problem 19"
m (→Solution 1) |
(→Video Solution) |
||
(15 intermediate revisions by 10 users not shown) | |||
Line 1: | Line 1: | ||
− | ==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> ? | ||
Line 5: | Line 5: | ||
==Solution 1== | ==Solution 1== | ||
− | Factoring out <math>98!+99!+100!</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>. 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 number of <math>5</math>’s in <math>98!(10,000)</math> is the same as the number of trailing zeros. Number of zeros is <math>98!</math> 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 number of <math>5</math>’s in <math>98!</math> which 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 == |
+ | https://youtu.be/alj9Y8jGNz8 | ||
+ | |||
+ | https://youtu.be/HISL2-N5NVg?t=817 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
+ | |||
+ | https://youtu.be/meEuDzrM5Ac | ||
+ | |||
+ | ~savannahsolver | ||
==See Also== | ==See Also== |
Latest revision as of 20:53, 9 March 2022
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 . Now has factors of , so there are a total of factors of .
Solution 2
Also keep in mind that number of ’s in is the same as the number of trailing zeros. Number of zeros is means we need pairs of ’s and ’s; we know there will be many more ’s, so we seek to find number of ’s in which solution tells us and that is factors of . has trailing zeros, so it has factors of and .
Video Solution
https://youtu.be/HISL2-N5NVg?t=817
~ pi_is_3.14
~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.