Difference between revisions of "2019 AMC 10A Problems/Problem 9"
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=== 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>. | 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>. | ||
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+ | === Solution 3 (Elimination) === | ||
+ | |||
+ | The sum of the first <math>n</math> positive integers is <math>\frac{(n)(n+1)}{2}</math> and the product of the positive integers upto <math>n</math> is <math>n!</math>. The quotient of the two is - | ||
+ | |||
+ | <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</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 == | ||
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~savannahsolver | ~savannahsolver | ||
− | ===Video Solution 2=== | + | ===Video Solution 2 by OmegaLearn=== |
https://youtu.be/FDgcLW4frg8?t=33 | https://youtu.be/FDgcLW4frg8?t=33 | ||
~ pi_is_3.14 | ~ pi_is_3.14 | ||
+ | |||
===Video Solution 3=== | ===Video Solution 3=== | ||
https://youtu.be/ikRv_0kNc2w | https://youtu.be/ikRv_0kNc2w |
Latest revision as of 16:31, 5 February 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 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
.
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.