Difference between revisions of "1989 AJHSME Problems/Problem 1"
Duoduoling0 (talk | contribs) (→Solution 2) |
Duoduoling0 (talk | contribs) (→Solution 2) |
||
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 &= \boxed{250}.</cmath> |
− | 10+30+50+70+90 &= \boxed{250}. | ||
~DuoDuoling0 | ~DuoDuoling0 |
Revision as of 12:21, 28 December 2021
Contents
[hide]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:
which gives us
\[10+30+50+70+90 &= \boxed{250}.\] (Error compiling LaTeX. Unknown error_msg)
~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.