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Difference between revisions of "2019 AMC 10B Problems/Problem 6"

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{{duplicate|[[2019 AMC 10B Problems|2019 AMC 10B #6]] and [[2019 AMC 12B Problems|2019 AMC 12B #4]]}}
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==Problem==
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There is a positive integer <math>n</math> such that <math>(n+1)! + (n+2)! = n! \cdot 440</math>. What is the sum of the digits of <math>n</math>?
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<math>\textbf{(A) }3\qquad\textbf{(B) }8\qquad\textbf{(C) }10\qquad\textbf{(D) }11\qquad\textbf{(E) }12</math>
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==Solution==
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===Solution 1===
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<cmath>\begin{split}& (n+1)n! + (n+2)(n+1)n! = 440 \cdot n! \\
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\Rightarrow \ &n![n+1 + (n+2)(n+1)] = 440 \cdot n! \\
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\Rightarrow \ &n + 1 + n^2 + 3n + 2 = 440 \\
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\Rightarrow \ &n^2 + 4n - 437 = 0\end{split}</cmath>
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Solving by the quadratic formula, <math>n = \frac{-4\pm \sqrt{16+437\cdot4}}{2} = \frac{-4\pm 42}{2} = \frac{38}{2} = 19</math> (since clearly <math>n \geq 0</math>). The answer is therefore <math>1 + 9 = \boxed{\textbf{(C) }10}</math>.
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===Solution 2===
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Dividing both sides by <math>n!</math> gives
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<cmath>(n+1)+(n+2)(n+1)=440 \Rightarrow n^2+4n-437=0 \Rightarrow (n-19)(n+23)=0.</cmath>
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Since <math>n</math> is non-negative, <math>n=19</math>. The answer is <math>1 + 9 = \boxed{\textbf{(C) }10}</math>.
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===Solution 3===
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Dividing both sides by <math>n!</math> as before gives <math>(n+1)+(n+1)(n+2)=440</math>. Now factor out <math>(n+1)</math>, giving <math>(n+1)(n+3)=440</math>. By considering the prime factorization of <math>440</math>, a bit of experimentation gives us <math>n+1=20</math> and <math>n+3=22</math>, so <math>n=19</math>, so the answer is <math>1 + 9 = \boxed{\textbf{(C) }10}</math>.
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===Solution 4===
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Since <math>(n+1)! + (n+2)! = (n+1)n! + (n+2)(n+1)n! = 440 \cdot n!</math>, the result can be factored into <math>(n+1)(n+3)n!=440 \cdot n!</math> and divided by <math>n!</math> on both sides to get <math>(n+1)(n+3)=440</math>. From there, it is easier to complete the square with the quadratic <math>(n+1)(n+3) = n^2 + 4n + 3</math>, so <math>n^2+4n+4=441 \Rightarrow (n+2)^2=441</math>. Solving for <math>n</math> results in <math>n=19,-23</math>, and since <math>n>0</math>, <math>n=19</math> and the answer is <math>1 + 9 = \boxed{\textbf{(C) }10}</math>.
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~Randomlygenerated
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===Solution 5===
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Rewrite <math>(n+1)! + (n+2)! = 440 \cdot n!</math> as <math>(n+1)n! + (n+2)(n+1)n! = 440 \cdot n!.</math> Factoring out the <math>n!</math> we get <math>n!(n + 1 + (n+1)(n+2)) = 440 \cdot n!.</math> Expand this to get <math>n!(n^2 + 4n + 3) = 440 \cdot n!.</math> Factor this and divide by <math>n!</math> to get <math>(n + 1)(n + 3) = 440.</math> If we take the prime factorization of <math>440</math> we see that it is <math>2^3 * 5 * 11.</math> Intuitively, we can find that <math>n + 1 = 20</math> and <math>n + 3 = 22.</math> Therefore, <math>n = 19.</math> Since the problem asks for the sum of the didgits of <math>n</math>, we finally calculate <math>1 + 9 = 10</math> and get answer choice <math>\boxed{\textbf{(C) }10}</math>.
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~pnacham
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==Video Solution==
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https://youtu.be/7p_ESPPF2es
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~Education, the Study of Everything
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== Video Solution==
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https://youtu.be/ba6w1OhXqOQ?t=1956
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==Video Solution==
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https://youtu.be/7xf_g3YQk00
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 +
~IceMatrix
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==Video Solution==
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https://youtu.be/6YFN_hwotUk
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~savannahsolver
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==See Also==
 +
 
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{{AMC10 box|year=2019|ab=B|num-b=5|num-a=7}}
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{{AMC12 box|year=2019|ab=B|num-b=3|num-a=5}}
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{{MAA Notice}}

Latest revision as of 09:24, 24 June 2023

The following problem is from both the 2019 AMC 10B #6 and 2019 AMC 12B #4, so both problems redirect to this page.

Problem

There is a positive integer $n$ such that $(n+1)! + (n+2)! = n! \cdot 440$. What is the sum of the digits of $n$?

$\textbf{(A) }3\qquad\textbf{(B) }8\qquad\textbf{(C) }10\qquad\textbf{(D) }11\qquad\textbf{(E) }12$

Solution

Solution 1

\[\begin{split}& (n+1)n! + (n+2)(n+1)n! = 440 \cdot n! \\ \Rightarrow \ &n![n+1 + (n+2)(n+1)] = 440 \cdot n! \\ \Rightarrow \ &n + 1 + n^2 + 3n + 2 = 440 \\ \Rightarrow \ &n^2 + 4n - 437 = 0\end{split}\]

Solving by the quadratic formula, $n = \frac{-4\pm \sqrt{16+437\cdot4}}{2} = \frac{-4\pm 42}{2} = \frac{38}{2} = 19$ (since clearly $n \geq 0$). The answer is therefore $1 + 9 = \boxed{\textbf{(C) }10}$.

Solution 2

Dividing both sides by $n!$ gives \[(n+1)+(n+2)(n+1)=440 \Rightarrow n^2+4n-437=0 \Rightarrow (n-19)(n+23)=0.\] Since $n$ is non-negative, $n=19$. The answer is $1 + 9 = \boxed{\textbf{(C) }10}$.

Solution 3

Dividing both sides by $n!$ as before gives $(n+1)+(n+1)(n+2)=440$. Now factor out $(n+1)$, giving $(n+1)(n+3)=440$. By considering the prime factorization of $440$, a bit of experimentation gives us $n+1=20$ and $n+3=22$, so $n=19$, so the answer is $1 + 9 = \boxed{\textbf{(C) }10}$.

Solution 4

Since $(n+1)! + (n+2)! = (n+1)n! + (n+2)(n+1)n! = 440 \cdot n!$, the result can be factored into $(n+1)(n+3)n!=440 \cdot n!$ and divided by $n!$ on both sides to get $(n+1)(n+3)=440$. From there, it is easier to complete the square with the quadratic $(n+1)(n+3) = n^2 + 4n + 3$, so $n^2+4n+4=441 \Rightarrow (n+2)^2=441$. Solving for $n$ results in $n=19,-23$, and since $n>0$, $n=19$ and the answer is $1 + 9 = \boxed{\textbf{(C) }10}$.

~Randomlygenerated

Solution 5

Rewrite $(n+1)! + (n+2)! = 440 \cdot n!$ as $(n+1)n! + (n+2)(n+1)n! = 440 \cdot n!.$ Factoring out the $n!$ we get $n!(n + 1 + (n+1)(n+2)) = 440 \cdot n!.$ Expand this to get $n!(n^2 + 4n + 3) = 440 \cdot n!.$ Factor this and divide by $n!$ to get $(n + 1)(n + 3) = 440.$ If we take the prime factorization of $440$ we see that it is $2^3 * 5 * 11.$ Intuitively, we can find that $n + 1 = 20$ and $n + 3 = 22.$ Therefore, $n = 19.$ Since the problem asks for the sum of the didgits of $n$, we finally calculate $1 + 9 = 10$ and get answer choice $\boxed{\textbf{(C) }10}$.

~pnacham

Video Solution

https://youtu.be/7p_ESPPF2es

~Education, the Study of Everything

Video Solution

https://youtu.be/ba6w1OhXqOQ?t=1956

Video Solution

https://youtu.be/7xf_g3YQk00

~IceMatrix

Video Solution

https://youtu.be/6YFN_hwotUk

~savannahsolver

See Also

2019 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
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 AMC 10 Problems and Solutions
2019 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
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 AMC 12 Problems and Solutions

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