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 | + | For any positive integer <math>M</math>, the notation M! denotes the product of the integers 1 through |
+ | M. 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) }327</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>. | + | 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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− | The | ||
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− | + | ~CHECKMATE2021 | |
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− | + | Note: Can you say what formula this uses? most AMC 8 test takers won't know it. Also, can someone unvandalize this page at the top? | |
+ | ==Video Solution (Omega Learn)== | ||
+ | https://www.youtube.com/watch?v=HISL2-N5NVg&t=817s | ||
− | + | ~ GeometryMystery | |
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==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}} |
Latest revision as of 14:47, 16 August 2024
wertesryrtutyrudtu
Problem
For any positive integer , the notation M! denotes the product of the integers 1 through
M. 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
Note: Can you say what formula this uses? most AMC 8 test takers won't know it. Also, can someone unvandalize this page at the top?
Video Solution (Omega Learn)
https://www.youtube.com/watch?v=HISL2-N5NVg&t=817s
~ GeometryMystery
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.