Difference between revisions of "1989 AJHSME Problems/Problem 1"
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&= 250 \Longrightarrow \boxed{\text{E}} | &= 250 \Longrightarrow \boxed{\text{E}} | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
| + | |||
| + | ==Solution 2== | ||
| + | Or, we can just combine the first, second, third, fourth, and last terms in each parentheses. This gives us: | ||
| + | <cmath>\begin{align*} | ||
| + | (1+9)+(11+19)+(21+29)+(31+39)+(41+49),</cmath> | ||
| + | which gives us | ||
| + | <math></math>\begin{align*} | ||
| + | <math>10+30+50+70+90 &= \boxed{250}.</math> | ||
==See Also== | ==See Also== | ||
Revision as of 12:20, 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:
\begin{align*}
(1+9)+(11+19)+(21+29)+(31+39)+(41+49), (Error compiling LaTeX. Unknown error_msg)
which gives us $$ (Error compiling LaTeX. Unknown error_msg)\begin{align*} $10+30+50+70+90 &= \boxed{250}.$ (Error compiling LaTeX. Unknown error_msg)
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. 
