Difference between revisions of "1970 AHSME Problems/Problem 22"

(Solution 2)
m (Solution 2)
 
Line 29: Line 29:
  
 
==Solution 2==
 
==Solution 2==
Expressing as an eqaution:
+
Expressing as an equation:
 
\begin{equation}\tag{1}\frac{3n(3n+1)}{2} = \frac{n(n+1)}{2} + 150.  \end{equation}
 
\begin{equation}\tag{1}\frac{3n(3n+1)}{2} = \frac{n(n+1)}{2} + 150.  \end{equation}
  

Latest revision as of 02:25, 27 April 2024

Problem

If the sum of the first $3n$ positive integers is $150$ more than the sum of the first $n$ positive integers, then the sum of the first $4n$ positive integers is

$\text{(A) } 300\quad \text{(B) } 350\quad \text{(C) } 400\quad \text{(D) } 450\quad \text{(E) } 600$

Solution 1

We can setup our first equation as

$\frac{3n(3n+1)}{2} = \frac{n(n+1)}{2} + 150$

Simplifying we get

$9n^2 + 3n = n^2 + n + 300 \Rightarrow 8n^2 + 2n - 300 = 0 \Rightarrow 4n^2 + n - 150 = 0$

So our roots using the quadratic formula are

$\dfrac{-b\pm\sqrt{b^2 - 4ac}}{2a} \Rightarrow \dfrac{-1\pm\sqrt{1^2 - 4\cdot(-150)\cdot4}}{2\cdot4} \Rightarrow \dfrac{-1\pm\sqrt{1+2400}}{8} \Rightarrow 6, -25/4$

Since the question said positive integers, $n = 6$, so $4n = 24$

$\frac{24\cdot 25}{2} = 300$

$\fbox{A}$

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)

$\frac{3n(3n+1)}{2} - \frac{n(n+1)}{2} = 150$

$=\frac{n(3(3n+1)-(n+1))}{2} = 150 =\frac{n(9n+3-n-1)}{2}$

$\frac{n(8n+2)}{2}= \frac{2n(4n+1)}{2}  = 150$

$\frac{4n(4n+1)}{2} = 2*150 = 300$

300 is the answer

$\fbox{A}$


〜Melkor

See also

1970 AHSC (ProblemsAnswer KeyResources)
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. AMC logo.png