Difference between revisions of "1985 AJHSME Problems/Problem 2"
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The expression is equal to the sum of integers from <math>1</math> to <math>99</math> minus the sum of integers from <math>1</math> to <math>89</math>, so it is equal to <math>\frac{99(100)}{2} - \frac{89(90)}{2} = 4950 - 4005 = \boxed{\text{(B)}~945}</math>. | The expression is equal to the sum of integers from <math>1</math> to <math>99</math> minus the sum of integers from <math>1</math> to <math>89</math>, so it is equal to <math>\frac{99(100)}{2} - \frac{89(90)}{2} = 4950 - 4005 = \boxed{\text{(B)}~945}</math>. | ||
| − | ~ cxsmi | + | ~ [https://artofproblemsolving.com/wiki/index.php/User:Cxsmi cxsmi] |
==Video Solution by BoundlessBrain!== | ==Video Solution by BoundlessBrain!== | ||
Revision as of 12:43, 4 April 2024
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:
![\[\frac{10}{2}\cdot(90+99)=\boxed{\text{(B)}~945}\]](http://latex.artofproblemsolving.com/9/e/7/9e7b8229c2e1de43dd8882831153b5f103b1ecf3.png)
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
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. 
