Difference between revisions of "2017 AMC 8 Problems/Problem 19"
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==Solution 3== | ==Solution 3== | ||
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>. | 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 is (2*5)^2</math> | ||
==Video Solution (CREATIVE THINKING + ANALYSIS!!!)== | ==Video Solution (CREATIVE THINKING + ANALYSIS!!!)== |
Revision as of 17:06, 21 January 2024
Contents
[hide]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.
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.