Difference between revisions of "1985 AJHSME Problems/Problem 2"
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We find a simpler problem in this problem, and simplify -> <math>90 + 91 + ... + 98 + 99 = 90 \times 10 + 1 + 2 + 3 + ... + 8 + 9</math> | We find a simpler problem in this problem, and simplify -> <math>90 + 91 + ... + 98 + 99 = 90 \times 10 + 1 + 2 + 3 + ... + 8 + 9</math> | ||
− | We know <math>90 \times 10</math>, that's easy | + | We know <math>90 \times 10</math>, that's easy: <math>900</math>. So how do we find <math>1 + 2 + ... + 8 + 9</math>? |
We rearrange the numbers to make <math>(1 + 9) + (2 + 8) + (3 + 7) + (4 + 6) + 5</math>. You might have noticed that each of the terms we put next to each other add up to 10, which makes for easy adding. <math>4 \times 10 + 5 = 45</math>. Adding that on to 900 makes 945. | We rearrange the numbers to make <math>(1 + 9) + (2 + 8) + (3 + 7) + (4 + 6) + 5</math>. You might have noticed that each of the terms we put next to each other add up to 10, which makes for easy adding. <math>4 \times 10 + 5 = 45</math>. Adding that on to 900 makes 945. |
Revision as of 15:22, 11 April 2020
Problem
Solution
Solution 1
One possibility is to simply add them. However, this can be time-consuming, and there are other ways to solve this problem.
We find a simpler problem in this problem, and simplify ->
We know , that's easy:
. So how do we find
?
We rearrange the numbers to make . You might have noticed that each of the terms we put next to each other add up to 10, which makes for easy adding.
. Adding that on to 900 makes 945.
945 is
Solution 2
Instead of breaking the sum and then rearranging, we can start by rearranging:
Solution 3
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:
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.