Difference between revisions of "2019 AMC 10A Problems/Problem 9"
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===Solution 1=== | ===Solution 1=== | ||
− | + | The sum of <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>, so would be able to cancel with these factors in <math>n!</math>, and thus the sum would be a divisor. 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 choice <math>D</math> does not work since <math>999</math> is divisible by <math>9</math>. Thus, the correct answer must be <math>\boxed{\textbf{(B) } 996}</math>. | |
==See Also== | ==See Also== |
Revision as of 00:33, 27 February 2019
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?
Solution 1
The sum of 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
, so would be able to cancel with these factors in
, and thus the sum would be a divisor. 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 choice
does not work since
is divisible by
. Thus, the correct answer must be
.
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.