Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 2)
m (→‎Solution 2)
 
(5 intermediate revisions by the same user not shown)
Line 29: Line 29:
  
 
==Solution 2==
 
==Solution 2==
−
Expressing as an eqaution:
+
Expressing as an equation:
−
\begin{equation}\frac{3n(3n+1)}{2} = \frac{n(n+1)}{2} + 150. \label{1} \end{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 =
 
The sum of the first 4n positive integers =
−
\begin{equation}\frac{4n(4n+1)}{2}\end{equation}
+
\begin{equation}\tag{2}\frac{4n(4n+1)}{2}\end{equation}
  
 
We will try to rearrange Equation (1) to give equation (2)
 
We will try to rearrange Equation (1) to give equation (2)
Line 41: Line 41:
 
<math>=\frac{n(3(3n+1)-(n+1))}{2} = 150 =\frac{n(9n+3-n-1)}{2}</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{n(8n+2)}{2}= \frac{2n(4n+1)}{2}  = 150</math>
  
−
<math>\frac{4n(4n+1}{2} = 2*150 = 300</math>
+
<math>\frac{4n(4n+1)}{2} = 2*150 = 300</math>
  
 
300 is the answer
 
300 is the answer

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 (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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1970_AHSME_Problems/Problem_22&oldid=218730"