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Difference between revisions of "2003 AIME I Problems/Problem 8"

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(→‎Solution 2)
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<cmath>d^2+3ad=30a</cmath>
 
<cmath>d^2+3ad=30a</cmath>
 
<cmath>d^2+3ad-30a=0</cmath>
 
<cmath>d^2+3ad-30a=0</cmath>
−
<cmath>d=\frac{-3a + \sqrt{9a^2+120a}}{2}</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  
+
Let <math>9a^2+120a=x^2</math>, where <math>x</math> is an integer. This yields the following:
 
<cmath>9a^2+120a-x^2=0</cmath>
 
<cmath>9a^2+120a-x^2=0</cmath>
−
<math></math>a=\frac{-120 \pm
+
<cmath>a=\frac{-120 + \sqrt{14400+36x^2}}{18}</cmath>
 +
<cmath>a=\frac{-20 + \sqrt{400+x^2}}{3}.</cmath>
 +
We then set <math>400+x^2=y^2</math>, where <math>y</math> is an integer. Factoring using difference of squares, we have
 +
<cmath>400=2^4 \cdot 5^2=(y+x)(y-x).</cmath>
 +
Then, noticing that <math>y+x > y-x</math>, we set up several systems of equations involving the factors of <math>400</math>. The first system we set up in this manner,
 +
<cmath>y+x=2^3 \cdot 5^2</cmath>
 +
<cmath>y-x=2,</cmath>
 +
yields the solution <math>y=101, x=99</math>. Plugging back in, we get that <math>a=27 \implies d=9</math>, so the sequence is <math>18,</math> <math>27,</math> <math>36,</math> <math>48,</math> and the answer is <math>\boxed{129}.</math>
 +
 
 +
*Note: we do not have to check the other systems since the <math>x</math> and <math>y</math> values obtained via this system yield integers for <math>a</math>, <math>d</math>, and this must be the only possible answer since this is an AIME problem. We also got very lucky that we didn't have to check more systems.
  
 
== See also ==
 
== See also ==

Revision as of 14:30, 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 the following: \[9a^2+120a-x^2=0\] \[a=\frac{-120 + \sqrt{14400+36x^2}}{18}\] \[a=\frac{-20 + \sqrt{400+x^2}}{3}.\] We then set $400+x^2=y^2$, where $y$ is an integer. Factoring using difference of squares, we have \[400=2^4 \cdot 5^2=(y+x)(y-x).\] Then, noticing that $y+x > y-x$, we set up several systems of equations involving the factors of $400$. The first system we set up in this manner, \[y+x=2^3 \cdot 5^2\] \[y-x=2,\] yields the solution $y=101, x=99$. Plugging back in, we get that $a=27 \implies d=9$, so the sequence is $18,$ $27,$ $36,$ $48,$ and the answer is $\boxed{129}.$

  • Note: we do not have to check the other systems since the $x$ and $y$ values obtained via this system yield integers for $a$, $d$, and this must be the only possible answer since this is an AIME problem. We also got very lucky that we didn't have to check more systems.

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

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