Difference between revisions of "1970 AHSME Problems/Problem 22"
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==Solution 2== | ==Solution 2== | ||
Expressing as an eqaution: | Expressing as an eqaution: | ||
| − | \begin{equation}\frac{3n(3n+1)}{2} = \frac{n(n+1)}{2} + 150 \end{equation} | + | \begin{equation}\frac{3n(3n+1)}{2} = \frac{n(n+1)}{2} + 150\label{eq:1} \end{equation} |
The sum of the first 4n positive integers =<math>\frac{4n(4n+1)}{2}</math>. | The sum of the first 4n positive integers =<math>\frac{4n(4n+1)}{2}</math>. | ||
Revision as of 02:17, 27 April 2024
Contents
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 1
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 
Solution 2
Expressing as an eqaution: \begin{equation}\frac{3n(3n+1)}{2} = \frac{n(n+1)}{2} + 150\label{eq:1} \end{equation}
The sum of the first 4n positive integers =
.
We will try to rearrange Equation (1) to give equation (2)
300 is the answer
〜Melkor
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. 
