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Difference between revisions of "1975 AHSME Problems/Problem 6"

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==Problem==
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The sum of the first eighty positive odd integers subtracted from the sum of the first eighty positive even integers is  
 
The sum of the first eighty positive odd integers subtracted from the sum of the first eighty positive even integers is  
  
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When the <math>n</math>th odd positive integer is subtracted from the <math>n</math>th even positive integer, the result is <math>1</math>. Therefore the sum of the first eighty positive odd integers subtracted from the sum of the first eighty positive even integers is <math>80\cdot1 = \boxed{\textbf{(E) } 80}</math>.
 
When the <math>n</math>th odd positive integer is subtracted from the <math>n</math>th even positive integer, the result is <math>1</math>. Therefore the sum of the first eighty positive odd integers subtracted from the sum of the first eighty positive even integers is <math>80\cdot1 = \boxed{\textbf{(E) } 80}</math>.
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==See Also==
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{{AHSME box|year=1975|num-b=5|num-a=7}}
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{{MAA Notice}}

Latest revision as of 15:52, 19 January 2021

Problem

The sum of the first eighty positive odd integers subtracted from the sum of the first eighty positive even integers is

$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 20 \qquad \textbf{(C)}\ 40 \qquad \textbf{(D)}\ 60 \qquad \textbf{(E)}\ 80$


Solution

Solution by e_power_pi_times_i


When the $n$th odd positive integer is subtracted from the $n$th even positive integer, the result is $1$. Therefore the sum of the first eighty positive odd integers subtracted from the sum of the first eighty positive even integers is $80\cdot1 = \boxed{\textbf{(E) } 80}$.

See Also

1975 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
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 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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