1985 AJHSME Problems/Problem 2
Contents
[hide]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: . , and .
Solution 2
We can express each of the terms as a difference from 100 and then add the negatives using to get the answer.
Solution 3
Instead of breaking the sum and then rearranging, we can start by rearranging:
Solution 4
We can use the formula for finite arithmetic sequences.
It is () where is the number of terms in the sequence, is the first term and is the last term.
Applying it here:
Video Solution
~savannahsolver
See Also
1985 AJHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
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All AJHSME/AMC 8 Problems and Solutions |
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