Difference between revisions of "1989 AJHSME Problems/Problem 1"
Duoduoling0 (talk | contribs) (→Solution) |
Duoduoling0 (talk | contribs) (→Solution 2) |
||
| (3 intermediate revisions by the same user not shown) | |||
| Line 15: | Line 15: | ||
==Solution 2== | ==Solution 2== | ||
Or, we can just combine the first, second, third, fourth, and last terms in each parentheses. This gives us: | Or, we can just combine the first, second, third, fourth, and last terms in each parentheses. This gives us: | ||
| − | <cmath> | + | |
| − | (1+9)+(11+19)+(21+29)+(31+39)+(41+49),</cmath> | + | <cmath>(1+9)+(11+19)+(21+29)+(31+39)+(41+49),</cmath> |
which gives us | which gives us | ||
| − | < | + | <cmath>10+30+50+70+90 = 250 = \boxed{E}.</cmath> |
| − | + | ||
| + | ~DuoDuoling0 | ||
==See Also== | ==See Also== | ||
Latest revision as of 12:21, 28 December 2021
Contents
Problem
Solution
We make use of the associative and commutative properties of addition to rearrange the sum as
Solution 2
Or, we can just combine the first, second, third, fourth, and last terms in each parentheses. This gives us:
![\[(1+9)+(11+19)+(21+29)+(31+39)+(41+49),\]](http://latex.artofproblemsolving.com/3/5/f/35fbf3b16902a4fab26c371d849b970cd4437d27.png)
which gives us
![\[10+30+50+70+90 = 250 = \boxed{E}.\]](http://latex.artofproblemsolving.com/4/c/8/4c8e39ae7fbe4e7352f6eee2b9ed18e3422cf201.png)
~DuoDuoling0
See Also
| 1989 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by First Problem |
Followed by Problem 2 | |
| 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. 
