Difference between revisions of "1985 AJHSME Problems/Problem 2"
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945 is <math>\boxed{\text{B}}</math> | 945 is <math>\boxed{\text{B}}</math> | ||
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We can express each of the terms as a difference from 100 and then add the negatives using <math>\frac{n(n+1)}{2}</math> to get the answer. | We can express each of the terms as a difference from 100 and then add the negatives using <math>\frac{n(n+1)}{2}</math> to get the answer. | ||
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
Revision as of 01:19, 10 June 2021
Problem
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
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:
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. 
