Difference between revisions of "1985 AJHSME Problems/Problem 2"
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===Solution 3=== | ===Solution 3=== | ||
| − | We can use a formula. It is n/2*(First term+Last term). Applying it here: | + | We can use a formula. |
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| + | It is <math>n/2*</math> (First term+Last term). Applying it here: | ||
<math>10/2 \times (90+99) = 945.</math> | <math>10/2 \times (90+99) = 945.</math> | ||
Revision as of 11:10, 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). 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 | ||
