Difference between revisions of "2019 AMC 10A Problems/Problem 9"
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<math>\textbf{(A) } 995 \qquad\textbf{(B) } 996 \qquad\textbf{(C) } 997 \qquad\textbf{(D) } 998 \qquad\textbf{(E) } 999</math> | <math>\textbf{(A) } 995 \qquad\textbf{(B) } 996 \qquad\textbf{(C) } 997 \qquad\textbf{(D) } 998 \qquad\textbf{(E) } 999</math> | ||
| − | == | + | ==Solutions== |
===Solution 1=== | ===Solution 1=== | ||
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Following from the fact that <math>n+1</math> must be prime, we can use to answer choices as possible solutions for <math>n</math>. <math>A</math>, <math>C</math>, and <math>E</math> don't work because <math>n+1</math> is even, and <math>D</math> does not work since <math>999</math> is divisible by <math>9</math>. Thus, the only correct answer is <math>996 \implies \boxed{\textbf{(B)}}</math>. | Following from the fact that <math>n+1</math> must be prime, we can use to answer choices as possible solutions for <math>n</math>. <math>A</math>, <math>C</math>, and <math>E</math> don't work because <math>n+1</math> is even, and <math>D</math> does not work since <math>999</math> is divisible by <math>9</math>. Thus, the only correct answer is <math>996 \implies \boxed{\textbf{(B)}}</math>. | ||
| + | |||
| + | ===Solution 3=== | ||
| + | There are only 5 numbers to test, so just test them to get B. | ||
==See Also== | ==See Also== | ||
Revision as of 22:33, 11 February 2019
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
Because the sum of positive integers is 
, and we want this to not be a divisor of the 
, 
must be prime. The greatest three-digit integer that is prime is 
. Subtract 
to get 
-Lcz
Solution 2
Following from the fact that must be prime, we can use to answer choices as possible solutions for . 
, 
, and 
don't work because is even, and 
does not work since 
is divisible by 
. Thus, the only correct answer is 
.
Solution 3
There are only 5 numbers to test, so just test them to get B.
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. 
