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 "2003 AIME I Problems/Problem 8"

(→‎Solution)
(→‎Solution)
Line 46: Line 46:
 
EDIT by NealShrestha:
 
EDIT by NealShrestha:
 
Note that once we reach <math>3ad + 4d^2 = 30a + 30d</math> this implies <math>3|d</math> since all other terms are congruent to <math>0\mod 3</math>.
 
Note that once we reach <math>3ad + 4d^2 = 30a + 30d</math> this implies <math>3|d</math> since all other terms are congruent to <math>0\mod 3</math>.
 +
 +
==Solution 2==
 +
The sequence is of the form <math>a-d,</math> <math>a,</math> <math>a+d,</math> <math>\frac{(a+d)^2}{a}</math>. Since the first and last terms differ by 30, we have <cmath>\frac{(a+d)^2}{a}-a+d=30</cmath>
 +
<cmath>d^2+3ad=30a</cmath>
 +
<cmath>d^2+3ad-30a=0</cmath>
 +
<cmath>d=\frac{-3a + \sqrt{9a^2+120a}}{2}</cmath>
 +
Let <math>9a^2+120a=x^2</math>, where <math>x</math> is an integer. This yields
 +
<cmath>9a^2+120a-x^2=0</cmath>
 +
<math></math>a=\frac{-120 \pm
  
 
== See also ==
 
== See also ==

Revision as of 13:14, 26 January 2022

Problem 8

In an increasing sequence of four positive integers, the first three terms form an arithmetic progression, the last three terms form a geometric progression, and the first and fourth terms differ by $30$. Find the sum of the four terms.

Solution

Denote the first term as $a$, and the common difference between the first three terms as $d$. The four numbers thus are in the form $a,\ a+d,\ a+2d,\ \frac{(a + 2d)^2}{a + d}$.

Since the first and fourth terms differ by $30$, we have that $\frac{(a + 2d)^2}{a + d} - a = 30$. Multiplying out by the denominator, \[(a^2 + 4ad + 4d^2) - a(a + d) = 30(a + d).\] This simplifies to $3ad + 4d^2 = 30a + 30d$, which upon rearranging yields $2d(2d - 15) = 3a(10 - d)$.

Both $a$ and $d$ are positive integers, so $2d - 15$ and $10 - d$ must have the same sign. Try if they are both positive (notice if they are both negative, then $d > 10$ and $d < \frac{15}{2}$, which is a contradiction). Then, $d = 8, 9$. Directly substituting and testing shows that $d \neq 8$, but that if $d = 9$ then $a = 18$. Alternatively, note that $3|2d$ or $3|2d-15$ implies that $3|d$, so only $9$ may work. Hence, the four terms are $18,\ 27,\ 36,\ 48$, which indeed fits the given conditions. Their sum is $\boxed{129}$.


Postscript

As another option, $3ad + 4d^2 = 30a + 30d$ could be rewritten as follows:


$d(3a + 4d) = 30(a + d)$


$d(3a + 3d)+ d^2 = 30(a + d)$


$3d(a + d)+ d^2 = 30(a + d)$


$(3d - 30)(a + d)+ d^2 = 0$


$3(d - 10)(a + d)+ d^2 = 0$


This gives another way to prove $d<10$, and when rewritten one last time:


$3(10 -d)(a + d) = d^2$


shows that $d$ must contain a factor of 3.


-jackshi2006

EDIT by NealShrestha: Note that once we reach $3ad + 4d^2 = 30a + 30d$ this implies $3|d$ since all other terms are congruent to $0\mod 3$.

Solution 2

The sequence is of the form $a-d,$ $a,$ $a+d,$ $\frac{(a+d)^2}{a}$. Since the first and last terms differ by 30, we have \[\frac{(a+d)^2}{a}-a+d=30\] \[d^2+3ad=30a\] \[d^2+3ad-30a=0\] \[d=\frac{-3a + \sqrt{9a^2+120a}}{2}\] Let $9a^2+120a=x^2$, where $x$ is an integer. This yields \[9a^2+120a-x^2=0\] $$ (Error compiling LaTeX. Unknown error_msg)a=\frac{-120 \pm

See also

2003 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME 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=2003_AIME_I_Problems/Problem_8&oldid=170285"