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

Revision as of 00:46, 19 October 2020

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 Solution

https://youtu.be/2vucE8HTiuU

~savannahsolver

@@ Line 1: / Line 1: @@
-==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?
 <math>\textbf{(A) } 995 \qquad\textbf{(B) } 996 \qquad\textbf{(C) } 997 \qquad\textbf{(D) } 998 \qquad\textbf{(E) } 999</math>
-===Solution 1===
+== Solutions ==
+=== Solution 1 ===
 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>.
-===Solution 2===
+=== Solution 2 ===
 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 Solution ===
 https://youtu.be/2vucE8HTiuU
 ~savannahsolver
-==See Also==
+== See Also ==
 {{AMC10 box|year=2019|ab=A|num-b=8|num-a=10}}
 {{MAA Notice}}

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