Problem:
1−2+3−4+5−6+7−8+910−9+8−7+6−5+4−3+2−1=
Answer Choices:
A. −1
B. 1
C. 5
D. 9
E. 10
Solution:
Group the numerator in pairs from the left, and group the denominator in pairs from the left:
Hence, the answer is 4(−1)+95(1)=55=1
OR
Regroup the numerator and denominator into positive and negative terms,
(1+3+5+7+9)−(2+4+6+8)(10+8+6+4+2)−(9+7+5+3+1)=25−2030−25=55=1.
Answer: B.
The problems on this page are the property of the MAA's American Mathematics Competitions
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