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Difference between revisions of "2019 AMC 10A Problems/Problem 9"
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==Problem==
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== Problem ==
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What is the greatest three-digit positive integer <math>n</math> for which the sum of the first <math>n</math> positive integers is <math>\underline{not}</math> a divisor of the product of the first <math>n</math> positive integers?
 
What is the greatest three-digit positive integer <math>n</math> for which the sum of the first <math>n</math> positive integers is <math>\underline{not}</math> a divisor of the product of the first <math>n</math> positive integers?
  
 
<math>\textbf{(A) } 995 \qquad\textbf{(B) } 996 \qquad\textbf{(C) } 997 \qquad\textbf{(D) } 998 \qquad\textbf{(E) } 999</math>
 
<math>\textbf{(A) } 995 \qquad\textbf{(B) } 996 \qquad\textbf{(C) } 997 \qquad\textbf{(D) } 998 \qquad\textbf{(E) } 999</math>
  
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===Solution 1===
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== Solutions ==
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The sum of the first <math>n</math> positive integers is <math>\frac{(n)(n+1)}{2}</math>, and we want this not to be a divisor of <math>n!</math> (the product of the first <math>n</math> positive integers). Notice that if and only if <math>n+1</math> were composite, all of its factors would be less than or equal to <math>n</math>, which means they would be able to cancel with the factors in <math>n!</math>. Thus, the sum of <math>n</math> positive integers would be a divisor of <math>n!</math> when <math>n+1</math> is composite. (Note: This is true for all positive integers except for 1 because 2 is not a divisor/factor of 1.) Hence in this case, <math>n+1</math> must instead be prime. The greatest three-digit integer that is prime is <math>997</math>, so we subtract <math>1</math> to get <math>n=\boxed{\textbf{(B) } 996}</math>.
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=== Solution 1 ===
 +
The sum of the first <math>n</math> positive integers is <math>\frac{(n)(n+1)}{2}</math>, and we want this to not be a divisor of <math>n!</math> (the product of the first <math>n</math> positive integers). Notice that if and only if <math>n+1</math> were composite, all of its factors would be less than or equal to <math>n</math>, which means they would be able to cancel with the factors in <math>n!</math>. Thus, the sum of <math>n</math> positive integers would be a divisor of <math>n!</math> when <math>n+1</math> is composite. (Note: This is true for all positive integers except for 1 because 2 is not a divisor/factor of 1.) Hence in this case, <math>n+1</math> must instead be prime. The greatest three-digit integer that is prime is <math>997</math>, so we subtract <math>1</math> to get <math>n=\boxed{\textbf{(B) } 996}</math>.
  
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===Solution 2===
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=== Solution 2 ===
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As in Solution 1, we deduce that <math>n+1</math> must be prime. If we can't immediately recall what the greatest three-digit prime is, we can instead use this result to eliminate answer choices as possible values of <math>n</math>. Choices <math>A</math>, <math>C</math>, and <math>E</math> don't work because <math>n+1</math> is even, and all even numbers are divisible by two, which makes choices <math>A</math>, <math>C</math>, and <math>E</math> composite and not prime. Choice <math>D</math> also does not work since <math>999</math> is divisible by <math>9</math>, which means it's a composite number and not prime. Thus, the correct answer must be <math>\boxed{\textbf{(B) } 996}</math>.
  
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As in Solution 1, we deduce that <math>n+1</math> must be prime. If we can't immediately recall what the greatest three-digit prime is, we can instead use this result to eliminate answer choices as possible values of <math>n</math>. Choices <math>A</math>, <math>C</math>, and <math>E</math> don't work because <math>n+1</math> is even, and all even numbers are divisible by two, which makes choices <math>A</math>, <math>C</math>, and <math>E</math> composite and not prime. Choice <math>D</math> also does not work since <math>999</math> is divisible by <math>9</math>, which means it's a composite number and not prime. Thus, the correct answer must be <math>\boxed{\textbf{(B) } 996}</math>.
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=== Solution 3 (Elimination) ===
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 +
The sum of the first <math>n</math> positive integers is <math>\frac{(n)(n+1)}{2}</math> and the product of the positive integers upto <math>n</math> is <math>n!</math>. The quotient of the two is -
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<math>\frac{(2)(n!)}{(n)(n+1)}</math>
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which simplifies to <math>\frac{(2)((n-1)!)}{n+1}</math>. Thus, <math>n+1</math> must be odd for the remainder to not be 0 (as <math>2</math> will multiply with some number in <math>n!</math>, cancelling out <math>n+1</math> if it is even, which leaves us with the answer choices <math>n = 996</math> and <math>n = 998</math>. Notice that <math>n + 1</math> must also be prime as otherwise there will be a factor of <math>n + 1</math> in <math>2</math> x <math>n!</math> somewhere. So either <math>997</math> or <math>999</math> must be prime - <math>999</math> is obviously not prime as it is divisible by 9, so our answer should be <math>n</math> where <math>n + 1 = 997</math>, and so our answer is <math>n = 996</math> or <math>\boxed{\textbf{(B) } 996}</math>.
 +
 
 +
- youtube.com/indianmathguy
  
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==Video Solution==
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== Video Solutions ==
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===Video Solution 1===
 
https://youtu.be/2vucE8HTiuU
 
https://youtu.be/2vucE8HTiuU
  
 
~savannahsolver
 
~savannahsolver
 +
===Video Solution 2 by OmegaLearn===
 +
https://youtu.be/FDgcLW4frg8?t=33
 +
 +
~ pi_is_3.14
 +
 +
===Video Solution 3===
 +
https://youtu.be/ikRv_0kNc2w
  
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==See Also==
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Education, the Study of Everything
  
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== See Also ==
 
{{AMC10 box|year=2019|ab=A|num-b=8|num-a=10}}
 
{{AMC10 box|year=2019|ab=A|num-b=8|num-a=10}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 09:41, 28 October 2024

Problem

What is the greatest three-digit positive integer $n$ for which the sum of the first $n$ positive integers is $\underline{not}$ a divisor of the product of the first $n$ positive integers?

$\textbf{(A) } 995 \qquad\textbf{(B) } 996 \qquad\textbf{(C) } 997 \qquad\textbf{(D) } 998 \qquad\textbf{(E) } 999$

Solutions

Solution 1

The sum of the first $n$ positive integers is $\frac{(n)(n+1)}{2}$, and we want this to not be a divisor of $n!$ (the product of the first $n$ positive integers). Notice that if and only if $n+1$ were composite, all of its factors would be less than or equal to $n$, which means they would be able to cancel with the factors in $n!$. Thus, the sum of $n$ positive integers would be a divisor of $n!$ when $n+1$ is composite. (Note: This is true for all positive integers except for 1 because 2 is not a divisor/factor of 1.) Hence in this case, $n+1$ must instead be prime. The greatest three-digit integer that is prime is $997$, so we subtract $1$ to get $n=\boxed{\textbf{(B) } 996}$.

Solution 2

As in Solution 1, we deduce that $n+1$ must be prime. If we can't immediately recall what the greatest three-digit prime is, we can instead use this result to eliminate answer choices as possible values of $n$. Choices $A$, $C$, and $E$ don't work because $n+1$ is even, and all even numbers are divisible by two, which makes choices $A$, $C$, and $E$ composite and not prime. Choice $D$ also does not work since $999$ is divisible by $9$, which means it's a composite number and not prime. Thus, the correct answer must be $\boxed{\textbf{(B) } 996}$.

Solution 3 (Elimination)

The sum of the first $n$ positive integers is $\frac{(n)(n+1)}{2}$ and the product of the positive integers upto $n$ is $n!$. The quotient of the two is -

$\frac{(2)(n!)}{(n)(n+1)}$

which simplifies to $\frac{(2)((n-1)!)}{n+1}$. Thus, $n+1$ must be odd for the remainder to not be 0 (as $2$ will multiply with some number in $n!$, cancelling out $n+1$ if it is even, which leaves us with the answer choices $n = 996$ and $n = 998$. Notice that $n + 1$ must also be prime as otherwise there will be a factor of $n + 1$ in $2$ x $n!$ somewhere. So either $997$ or $999$ must be prime - $999$ is obviously not prime as it is divisible by 9, so our answer should be $n$ where $n + 1 = 997$, and so our answer is $n = 996$ or $\boxed{\textbf{(B) } 996}$.

- youtube.com/indianmathguy

Video Solutions

Video Solution 1

https://youtu.be/2vucE8HTiuU

~savannahsolver

Video Solution 2 by OmegaLearn

https://youtu.be/FDgcLW4frg8?t=33

~ pi_is_3.14

Video Solution 3

https://youtu.be/ikRv_0kNc2w

Education, the Study of Everything

See Also

2019 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
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All AMC 10 Problems and Solutions

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