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

m (Solution 2)
m (Solution 1)
Line 6: Line 6:
  
 
===Solution 1===
 
===Solution 1===
The sum of <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>, so would be able to cancel with these factors in <math>n!</math>, and thus the sum would be a divisor. 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>.
+
The sum of <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. 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>.
  
 
===Solution 2===
 
===Solution 2===

Revision as of 18:28, 29 December 2019

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?

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

Solution 1

The sum of $n$ 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. 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 $\boxed{\textbf{(B) } 996}$.

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS