Difference between revisions of "1985 AJHSME Problems/Problem 2"
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<math>\frac{10}{2} \times (90+99) = 945 \rightarrow \boxed{B}</math> | <math>\frac{10}{2} \times (90+99) = 945 \rightarrow \boxed{B}</math> | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/1NtsgKc6mXs | ||
| + | |||
| + | ~savannahsolver | ||
==See Also== | ==See Also== | ||
Revision as of 07:37, 7 January 2023
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:
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
