Difference between revisions of "2019 AMC 10A Problems/Problem 9"

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=== Video Solution ===
 
 
https://youtu.be/ikRv_0kNc2w
 
 
Education, the Study of Everything
 
 
 
== Problem ==
 
== Problem ==
 
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?
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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>.
 
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>.
  
=== Video Solution ===
+
== Video Solutions ==
 +
===Video Solution 1===
 
https://youtu.be/2vucE8HTiuU
 
https://youtu.be/2vucE8HTiuU
  
 
~savannahsolver
 
~savannahsolver
 
+
===Video Solution 2===
== Video Solution ==
 
 
https://youtu.be/FDgcLW4frg8?t=33
 
https://youtu.be/FDgcLW4frg8?t=33
  
 
~ pi_is_3.14
 
~ pi_is_3.14
 +
===Video Solution 3===
 +
https://youtu.be/ikRv_0kNc2w
 +
 +
Education, the Study of Everything
  
 
== See Also ==
 
== 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}}

Revision as of 18:53, 13 April 2021

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 not to 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 $\boxed{\textbf{(B) } 996}$.

Video Solutions

Video Solution 1

https://youtu.be/2vucE8HTiuU

~savannahsolver

Video Solution 2

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 (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
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

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