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

Revision as of 19:00, 20 April 2021

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 $5$ . Now $10,000$ has $4$ factors of $5$ , so there are a total of $22 + 4 = \boxed{\textbf{(D)}\ 26}$ factors of $5$ .

Solution 2

The number of $5$ 's in the factorization of $98! + 99! + 100!$ is the same as the number of trailing zeroes. The number of zeroes is taken by the floor value of each number divided by $5$ , until you can't divide by $5$ anymore. Factorizing $98! + 99! + 100!$ , you get $98!(1+99+9900)=98!(10000)$ . To find the number of trailing zeroes in 98!, we do $\left\lfloor\frac{98}{5}\right\rfloor + \left\lfloor\frac{19}{5}\right\rfloor= 19 + 3=22$ . Now since $10000$ has 4 zeroes, we add $22 + 4$ to get $\boxed{\textbf{(D)}\ 26}$ factors of $5$ .

Video Solution

https://youtu.be/HISL2-N5NVg?t=817

~ pi_is_3.14

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.

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 ==Solution 2==
 The number of <math>5</math>'s in the factorization of <math>98! + 99! + 100!</math> is the same as the number of trailing zeroes. The number of zeroes is taken by the floor value of each number divided by <math>5</math>, until you can't divide by <math>5</math> anymore. Factorizing <math>98! + 99! + 100!</math>, you get <math>98!(1+99+9900)=98!(10000)</math>. To find the number of trailing zeroes in 98!, we do <math>\left\lfloor\frac{98}{5}\right\rfloor + \left\lfloor\frac{19}{5}\right\rfloor= 19 + 3=22</math>. Now since <math>10000</math> has 4 zeroes, we add <math>22 + 4</math>  to get <math>\boxed{\textbf{(D)}\ 26}</math> factors of <math>5</math>.
-==Solution 3==
-We can rewrite the expression as <math>98!+99!+100!=98!(1+99+99\cdot100)=98!(100+99\cdot100)=98!\cdot10,000</math>.
-The exponent of <math>5</math> in <math>10,000</math> is <math>4</math>. Onto the <math>98!</math> part.
-Remember that <math>98!</math> is the product of the integers from 1 to 98. Among these, there are multiples of <math>5</math> and multiples of <math>25</math>.
-The number of multiples of <math>5</math> below or equal to <math>98</math> is <math>\left\lfloor\frac{98}{5}\right\rfloor</math> (try to see why), which is <math>19</math>. Every such number has a factor of <math>5</math>, so they contribute <math>1</math> to the total each. So these numbers contribute <math>19</math> to the exponent of <math>5</math> in <math>98!</math>.
-However, we forgot about multiples of <math>25</math>. Using similar logic, we have <math>\left\lfloor\frac{98}{25}\right\rfloor=3</math>, so there are 3 multiples of 25 in this range. Multiples of 25 contribute 2 to the total each. We already counted 1 of 2 contributions while counting multiples of 5. So we need to add another 1 to the exponent of 5 for every multiple of 25. So these numbers contribute 3 to the exponent of 5 in <math>98!</math>.
-We thus have that the exponent of <math>5</math> in <math>98!</math> is <math>19 + 3 = 22</math>, and so our answer is <math>22 + 4 = 26 \longrightarrow \boxed{(D)26}</math>.
-~Math4Life2020
 == Video Solution ==

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