Difference between revisions of "1985 AJHSME Problems/Problem 2"
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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. | ||
| − | 945 is <math>\boxed{\text{B}}</math> | + | 945 is <math>\boxed{\text{B}}.</math> |
===Solution 2=== | ===Solution 2=== | ||
Revision as of 19:46, 6 May 2018
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 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
