Difference between revisions of "1975 AHSME Problems/Problem 6"
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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== | ||
| + | {{AHSME box|year=1975|num-b=5|num-a=7}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 16:52, 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
Solution
Solution by e_power_pi_times_i
When the 
th odd positive integer is subtracted from the th even positive integer, the result is 
. Therefore the sum of the first eighty positive odd integers subtracted from the sum of the first eighty positive even integers is 
.
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
