Difference between revisions of "2017 AMC 8 Problems/Problem 19"

Latest revision as of 03:20, 13 October 2024

Problem

For any positive integer $M$ , the notation $M!$ denotes the product of the integers $1$ through $M$ . What is the largest integer $n$ for which $5^n$ is a factor of the sum $98!+99!+100!$ ?

$\textbf{(A) }23 \qquad \textbf{(B) }24 \qquad \textbf{(C) }25 \qquad \textbf{(D) }26 \qquad \textbf{(E) }27$

Solution 1

Factoring out $98!+99!+100!$ , we have $98! (1+99+99*100)$ , which is $98! (10000)$ . Next, $98!$ has $\left\lfloor\frac{98}{5}\right\rfloor + \left\lfloor\frac{98}{25}\right\rfloor = 19 + 3 = 22$ factors of $5$ . The $19$ is because of all the multiples of $5$ .The $3$ is because of all the multiples of $25$ . Now, $10,000$ has $4$ factors of $5$ , so there are a total of $22 + 4 = \boxed{\textbf{(D)}\ 26}$ factors of $5$ .

~CHECKMATE2021

Note: Can you say what formula this uses? most AMC 8 test takers won't know it.

Video Solution (Omega Learn)

https://www.youtube.com/watch?v=HISL2-N5NVg&t=817s

~ GeometryMystery

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.

@@ Line 1: / Line 1: @@
 ==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> ?
-<math>\textbf{(A) }23\qquad\textbf{(B) }24\qquad\textbf{(C) }25\qquad\textbf{(D) }26\qquad\textbf{(E) }27</math>
+<math>\textbf{(A) }23 \qquad \textbf{(B) }24 \qquad \textbf{(C) }25 \qquad \textbf{(D) }26 \qquad \textbf{(E) }27</math>
 ==Solution 1==
@@ Line 9: / Line 10: @@
 ~CHECKMATE2021
-==Solution 2==
+Note: Can you say what formula this uses? most AMC 8 test takers won't know it.
-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 = \boxed{26}</math>.
-~CHECKMATE2021
+==Video Solution (Omega Learn)==
+https://www.youtube.com/watch?v=HISL2-N5NVg&t=817s
-==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>.
-==Video Solution (CREATIVE THINKING + ANALYSIS!!!)==
-https://youtu.be/WKux87BEO1U
-~Education, the Study of Everything
-== 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
+~ GeometryMystery
 ==See Also==

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Difference between revisions of "2017 AMC 8 Problems/Problem 19"

Latest revision as of 03:20, 13 October 2024

Contents

Problem

Solution 1

Video Solution (Omega Learn)

See Also