Difference between revisions of "2017 AMC 8 Problems/Problem 19"
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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>. | ||
− | Let's first find how many factors of <math>5 10,000</math> has. <math>10,000 is (2*5)^2</math> | + | 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> |
+ | |||
+ | 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> | ||
+ | |||
+ | 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> | ||
+ | |||
+ | 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> | ||
+ | |||
+ | So the answer is <math>26</math>. Which is answer choice D <math>\boxed{\textbf{(D)}\ 26}</math>. | ||
+ | |||
+ | ~CHECKMATE2021 | ||
==Video Solution (CREATIVE THINKING + ANALYSIS!!!)== | ==Video Solution (CREATIVE THINKING + ANALYSIS!!!)== |
Revision as of 17:18, 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.
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.