Difference between revisions of "1970 AHSME Problems/Problem 22"
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\text{(E) } 600</math> | \text{(E) } 600</math> | ||
| − | == Solution == | + | == Solution 1== |
| + | We can setup our first equation as | ||
| + | |||
| + | <math>\frac{3n(3n+1)}{2} = \frac{n(n+1)}{2} + 150</math> | ||
| + | |||
| + | Simplifying we get | ||
| + | |||
| + | <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 | ||
| + | |||
| + | <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> | ||
| + | |||
| + | 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> | ||
| + | |||
| + | ==Solution 2== | ||
| + | Expressing as an equation: | ||
| + | \begin{equation}\tag{1}\frac{3n(3n+1)}{2} = \frac{n(n+1)}{2} + 150. \end{equation} | ||
| + | |||
| + | The sum of the first 4n positive integers = | ||
| + | \begin{equation}\tag{2}\frac{4n(4n+1)}{2}\end{equation} | ||
| + | |||
| + | We will try to rearrange Equation (1) to give equation (2) | ||
| + | |||
| + | <math>\frac{3n(3n+1)}{2} - \frac{n(n+1)}{2} = 150</math> | ||
| + | |||
| + | <math>=\frac{n(3(3n+1)-(n+1))}{2} = 150 =\frac{n(9n+3-n-1)}{2}</math> | ||
| + | |||
| + | <math>\frac{n(8n+2)}{2}= \frac{2n(4n+1)}{2} = 150</math> | ||
| + | |||
| + | <math>\frac{4n(4n+1)}{2} = 2*150 = 300</math> | ||
| + | |||
| + | 300 is the answer | ||
| + | |||
<math>\fbox{A}</math> | <math>\fbox{A}</math> | ||
| + | |||
| + | |||
| + | 〜Melkor | ||
== See also == | == See also == | ||
| − | {{AHSME box|year=1970|num-b=21|num-a=23}} | + | {{AHSME 35p box|year=1970|num-b=21|num-a=23}} |
[[Category: Introductory Algebra Problems]] | [[Category: Introductory Algebra Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 02:25, 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 equation: \begin{equation}\tag{1}\frac{3n(3n+1)}{2} = \frac{n(n+1)}{2} + 150. \end{equation}
The sum of the first 4n positive integers = \begin{equation}\tag{2}\frac{4n(4n+1)}{2}\end{equation}
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. 
