Difference between revisions of "1985 AJHSME Problems/Problem 2"
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90+91+92+\cdots +98+99 &= (90+99)+(91+98)+(92+97)+(93+96)+(94+95) \\ | 90+91+92+\cdots +98+99 &= (90+99)+(91+98)+(92+97)+(93+96)+(94+95) \\ | ||
&= 189+189+189+189+189 \\ | &= 189+189+189+189+189 \\ | ||
| − | &= 945\rightarrow \boxed{\text{ | + | &= 945\rightarrow \boxed{\text{B}} |
\end{align*}</cmath> | \end{align*}</cmath> | ||
Revision as of 11:13, 20 April 2010
Problem
Solution
Solution 1
We could just add them all together. But what would be the point of doing that? So we find a slicker way.
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 a formula.
It is 
(First term+Last term) where 
is the number of terms in the sequence.
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 | ||
