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

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<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>
 
==Solutions==
 
  
 
===Solution 1===
 
===Solution 1===
Because the sum of <math>n</math> positive integers is <math>\frac{(n)(n+1)}{2}</math>, and we want this to not be a divisor of the <math>n!</math>, <math>n+1</math> must be prime. The greatest three-digit integer that is prime is <math>997</math>. Subtract <math>1</math> to get <math>996 \implies \boxed{\textbf{(B)}}.</math>
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Because the sum of <math>n</math> positive integers is <math>\frac{(n)(n+1)}{2}</math>, and we want this to not be a divisor of the <math>n!</math>, <math>n+1</math> must be prime. The greatest three-digit integer that is prime is <math>997</math>. Subtract <math>1</math> to get <math>\boxed{\textbf{(B) } 996}</math>.
 
 
-Lcz
 
  
 
===Solution 2===
 
===Solution 2===
  
We can use the fact that <math>n+1</math> must be prime to eliminate answer choices as possible values of <math>n</math>. <math>A</math>, <math>C</math>, and <math>E</math> don't work because <math>n+1</math> is even, and <math>D</math> does not work since <math>999</math> is divisible by <math>9</math>. Thus, the only correct answer is <math>996 \implies \boxed{\textbf{(B)}}</math>.
+
We can use the fact that <math>n+1</math> must be prime to eliminate answer choices as possible values of <math>n</math>. <math>A</math>, <math>C</math>, and <math>E</math> don't work because <math>n+1</math> is even, and <math>D</math> does not work since <math>999</math> is divisible by <math>9</math>. Thus, the only correct answer is <math>\boxed{\textbf{(B) } 996}</math>.
  
 
==See Also==
 
==See Also==

Revision as of 21:09, 17 February 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

Because the sum of $n$ positive integers is $\frac{(n)(n+1)}{2}$, and we want this to not be a divisor of the $n!$, $n+1$ must be prime. The greatest three-digit integer that is prime is $997$. Subtract $1$ to get $\boxed{\textbf{(B) } 996}$.

Solution 2

We can use the fact that $n+1$ must be prime to eliminate answer choices as possible values of $n$. $A$, $C$, and $E$ don't work because $n+1$ is even, and $D$ does not work since $999$ is divisible by $9$. Thus, the only correct answer is $\boxed{\textbf{(B) } 996}$.

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

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