Difference between revisions of "1970 AHSME Problems/Problem 22"
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== Solution == | == Solution == | ||
| + | We can setup our first equation as | ||
| + | |||
| + | <math>\frac{3n(3n+1)}{2} = \frac{n(n+1)}{2} + 150</math> | ||
| + | |||
| + | Simplifying we get | ||
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| + | <math>9n^2 + 3n = n^2 + n + 300 \Rightarrow 8n^2 + 2n - 300 = 0 \Rightarrow 4n^2 + n - 150 = 0</math> | ||
| + | |||
| + | So our roots using the quadratic formula are | ||
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| + | <math>\dfrac{-b\pm\sqrt{b^2 - 4ac}}{2a} \Rightarrow \dfrac{-1\pm\sqrt{1^2 - 4\cdot(-150)\cdot4}}{2\cdot4} \Rightarrow \dfrac{-1\pm\sqrt{1+2400}}{8} \Rightarrow 6, -25/4</math> | ||
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| + | Since the question said positive integers, <math> n = 6</math>, so <math>4n = 24</math> | ||
| + | |||
| + | <math>\frac{24\cdot 25}{2} = 300</math> | ||
| + | |||
<math>\fbox{A}</math> | <math>\fbox{A}</math> | ||
Latest revision as of 17:18, 27 November 2016
Problem
If the sum of the first 
positive integers is 
more than the sum of the first 
positive integers, then the sum of the first 
positive integers is
Solution
We can setup our first equation as
Simplifying we get
So our roots using the quadratic formula are
Since the question said positive integers, 
, so 
See also
| 1970 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 21 |
Followed by Problem 23 | |
| 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 • 31 • 32 • 33 • 34 • 35 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
