Difference between revisions of "2019 AMC 10A Problems/Problem 9"
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== Solutions == | == Solutions == | ||
=== Solution 1 === | === 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 | + | The sum of the first <math>n</math> positive integers is <math>\frac{(n)(n+1)}{2}</math>, and we want this to not 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 === | ||
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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</math> x <math>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</math> x <math>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>. | ||
| − | - | + | - youtube.com/indianmathguy |
== Video Solutions == | == Video Solutions == | ||
Latest revision as of 09:41, 28 October 2024
Contents
Problem
What is the greatest three-digit positive integer 
for which the sum of the first positive integers is 
a divisor of the product of the first positive integers?
Solutions
Solution 1
The sum of the first positive integers is 
, and we want this to not be a divisor of 
(the product of the first positive integers). Notice that if and only if 
were composite, all of its factors would be less than or equal to , which means they would be able to cancel with the factors in . Thus, the sum of positive integers would be a divisor of when 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, must instead be prime. The greatest three-digit integer that is prime is 
, so we subtract 
to get 
.
Solution 2
As in Solution 1, we deduce that 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 . Choices 
, 
, and 
don't work because is even, and all even numbers are divisible by two, which makes choices , , and composite and not prime. Choice 
also does not work since 
is divisible by 
, which means it's a composite number and not prime. Thus, the correct answer must be 
.
Solution 3 (Elimination)
The sum of the first positive integers is and the product of the positive integers upto is . The quotient of the two is -
which simplifies to 
. Thus, must be odd for the remainder to not be 0 (as 
will multiply with some number in , cancelling out if it is even, which leaves us with the answer choices 
and 
. Notice that 
must also be prime as otherwise there will be a factor of in x somewhere. So either or must be prime - is obviously not prime as it is divisible by 9, so our answer should be where 
, and so our answer is or .
- youtube.com/indianmathguy
Video Solutions
Video Solution 1
~savannahsolver
Video Solution 2 by OmegaLearn
https://youtu.be/FDgcLW4frg8?t=33
~ pi_is_3.14
Video Solution 3
Education, the Study of Everything
See Also
| 2019 AMC 10A (Problems • Answer Key • Resources) | ||
| 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. 
