1985 AJHSME Problems/Problem 2
Contents
Problem
Solution 1
To simplify the problem, we can group 90’s together: [mathjax]90 + 91 + ... + 98 + 99 = 90 \cdot 10 + 1 + 2 + 3 + ... + 8 + 9[/mathjax].
[mathjax]90\cdot10=900[/mathjax], and finding [mathjax]1 + 2 + ... + 8 + 9[/mathjax] has a trick to it.
Rearranging the numbers so each pair sums up to 10, we have: [mathjax display=true](1 + 9)+(2+8)+(3+7)+(4+6)+5[/mathjax]. [mathjax]4\cdot10+5 = 45[/mathjax], and [mathjax]900+45=\boxed{\text{(B)}~945}[/mathjax].
Solution 2
We can express each of the terms as a difference from and then add the negatives using to get the answer.
Solution 3
Instead of breaking the sum then rearranging, we can rearrange directly:
Solution 4
The finite arithmetic sequence formula states that the sum in the sequence is equal to where is the number of terms in the sequence, is the first term and is the last term.
Applying the formula, we have:
Solution 5
The expression is equal to the sum of integers from to minus the sum of integers from to , so it is equal to .
~ cxsmi
Solution 6 (Estimate)
Notice the sum is larger than and less than
The only choice that works is 945, thus the answer is
~ lovelearning999
Solution 7 (Brute Force)
Add all the numbers to get 945, or
~ lovelearning999
Video Solution by BoundlessBrain!
Video Solution
~savannahsolver
See Also
1985 AJHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
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 AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.