Difference between revisions of "1989 AJHSME Problems/Problem 1"
5849206328x (talk | contribs) (New page: ==Problem== <math>(1+11+21+31+41)+(9+19+29+39+49)=</math> <math>\text{(A)}\ 150 \qquad \text{(B)}\ 199 \qquad \text{(C)}\ 200 \qquad \text{(D)}\ 249 \qquad \text{(E)}\ 250</math> ==Solu...) |
Duoduoling0 (talk | contribs) (→Solution 2) |
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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>(1+9)+(11+19)+(21+29)+(31+39)+(41+49),</cmath> | ||
+ | which gives us | ||
+ | <cmath>10+30+50+70+90 = 250 = \boxed{E}.</cmath> | ||
+ | |||
+ | ~DuoDuoling0 | ||
==See Also== | ==See Also== | ||
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{{AJHSME box|year=1989|before=First<br>Problem|num-a=2}} | {{AJHSME box|year=1989|before=First<br>Problem|num-a=2}} | ||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
+ | {{MAA Notice}} |
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:
which gives us
~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.