Difference between revisions of "1985 AJHSME Problems/Problem 2"
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Applying the formula, we have: | Applying the formula, we have: | ||
<cmath>\frac{10}{2}\cdot(90+99)=\boxed{\text{(B)}~945}</cmath> | <cmath>\frac{10}{2}\cdot(90+99)=\boxed{\text{(B)}~945}</cmath> | ||
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| + | ==Video Solution by BoundlessBrain! | ||
==Video Solution== | ==Video Solution== | ||
Revision as of 09:50, 24 June 2023
Problem
Solution 1
To simplify the problem, we can group 90’s together: 
.
, and finding 
has a trick to it.
Rearranging the numbers so each pair sums up to 10, we have:
![\[(1 + 9)+(2+8)+(3+7)+(4+6)+5\]](http://latex.artofproblemsolving.com/1/5/1/151425ffc3700df99ccddfbfcf3c4452458051d3.png)
. 
, and 
.
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)
==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. 
