Difference between revisions of "2019 AMC 10A Problems/Problem 9"
(→Problem) |
m (Compilation of all Video Solutions) |
||
| Line 17: | Line 17: | ||
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>. | 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 Solutions == |
| + | ===Video Solution 1=== | ||
https://youtu.be/2vucE8HTiuU | https://youtu.be/2vucE8HTiuU | ||
~savannahsolver | ~savannahsolver | ||
| + | ===Video Solution 2=== | ||
| + | https://youtu.be/FDgcLW4frg8?t=33 | ||
| + | |||
| + | ~ pi_is_3.14 | ||
| + | ===Video Solution 3=== | ||
| + | https://youtu.be/ikRv_0kNc2w | ||
| + | |||
| + | Education, the Study of Everything | ||
== Video Solution == | == Video Solution == | ||
Revision as of 17:53, 13 April 2021
Contents
Video Solution
Education, the Study of Everything
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 not to 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 
.
Video Solutions
Video Solution 1
~savannahsolver
Video Solution 2
https://youtu.be/FDgcLW4frg8?t=33
~ pi_is_3.14
Video Solution 3
Education, the Study of Everything
Video Solution
https://youtu.be/FDgcLW4frg8?t=33
~ pi_is_3.14
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. 
