1985 AJHSME Problems/Problem 2
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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