1990 AJHSME Problems/Problem 16
Contents
Problem
Solution
In the middle, we have 
.
If we match up the back with the front, and then do the same for the rest, we get pairs with 
and 
, so these will cancel out. In the middle, we have 
which doesn't cancel, but gives us 
.
Solution 2
We can see that there are 
terms in total. We can also see that the first 
numbers form groups of two that add to 
each. Dividing to see how many pairs we have, /
= 
groups of ten, or 
. However, we have to remember to add the 199th term (), so we get + = 
, which gives us 
.
See Also
| 1990 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 15 |
Followed by Problem 17 | |
| 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 | ||
