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

Revision as of 13:08, 18 January 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 $5$ .

Solution 3

We can rewrite the expression as $98!+99!+100!=98!(1+99+99\cdot100)=98!(100+99\cdot100)=98!\cdot10,000$ . The exponent of $5$ in $10,000$ is $4$ . Onto the $98!$ part.

Remember that $98!$ is the product of the integers from 1 to 98. Among these, there are multiples of $5$ and multiples of $25$ . The number of multiples of $5$ below or equal to $98$ is $\left\lfloor\frac{98}{5}\right\rfloor$ (try to see why), which is $19$ . Every such number has a factor of $5$ , so they contribute $1$ to the total each. So these numbers contribute $19$ to the exponent of $5$ in $98!$ .

However, we forgot about multiples of $25$ . Using similar logic, we have $\left\lfloor\frac{98}{25}\right\rfloor=3$ , 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 $98!$ .

We thus have that the exponent of $5$ in $98!$ is $19 + 3 = 22$ , and so our answer is $22 + 4 = 26 \longrightarrow \boxed{(D)26}$ .

~Math4Life2020

Revision as of 13:37, 26 December 2020 (view source) Math4life2020 (talk \| contribs) (Added another solution.) ← Older edit		Revision as of 13:08, 18 January 2021 (view source) Hashtagmath (talk \| contribs) Newer edit →
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−	==Problem 19==	+	==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> ?

2017 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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