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

Revision as of 03:07, 5 November 2023

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 $\boxed{\textbf{(B) } 996}$ .

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 $2 * n!$ 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 $\boxed{\textbf{(B) } 996}$ .

- abed_nadir

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

@@ Line 17: / Line 17: @@
 <math>\frac{(2)(n!)}{(n)(n+1)}</math>
-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 x 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>.
+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 * 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>.
 - abed_nadir

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