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

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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>
 
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.
+
*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 got very lucky in this sense :)
 +
-Fasolinka
  
 
== See also ==
 
== See also ==

Revision as of 15:37, 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 got very lucky in this sense :)

-Fasolinka

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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