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

Latest revision as of 22:21, 22 January 2023

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 second 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

Solution 3 (Guesswork)

We represent the values as $a-d$ , $a$ , $a+d$ , and $\frac{(a+d)^2}{a}$ Take the difference between the first and last values $\frac{(a+d)^2}{a}-a+d=30$ Manipulating the values by expanding and then long division we see $\frac{a^2+2ad+d^2}{a}-a+d=30$ $\frac{(a+2d)a+d^2}{a}-a+d=30$ $a+2d+\frac{d^2}{a}-a+d=30$ Combining like terms we get $3d+\frac{d^2}{a}=30$ And looking at the values we know that since the second term must be positive (since both a and d are positive), $d$ must be a maximum of 9, which offers 9 possible values (since $d$ must be a positive integer.) We can resort to guesswork by this time, and thus receive the result of of $d=9$ at which $a=27$ . The solution is thus $\boxed{129}.$

We could also lay down some further parameters for the value of $d$ . as $a$ must be greater than $d$ (for the first term to be positive), we can substitite in a "borderline" value of a=d (which itself is just over the limit). This gives us:

$3d+\frac{d^2}{d}=30$ $3d+d=30$ $4d=30$ Thus, d must be greater than 7.5, which gives us only 2 values left (8 and 9). We can then plug in any of the two to see if an integer value is attainable. In the end, 8 doesn't work and 9 does, which gives the solution of $\boxed{129}.$

Note: This solution is similar to Solution 2

-PMaldonado13

@@ Line 43: / Line 43: @@
 -jackshi2006
 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>.
+==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 the following:
+<cmath>9a^2+120a-x^2=0</cmath>
+<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 second 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 got very lucky in this sense :)
+-Fasolinka
+==Solution 3 (Guesswork)==
+We represent the values as <math>a-d</math>, <math>a</math>, <math>a+d</math>, and <math>\frac{(a+d)^2}{a}</math>
+Take the difference between the first and last values
+<cmath>\frac{(a+d)^2}{a}-a+d=30</cmath>
+Manipulating the values by expanding and then long division we see
+<cmath>\frac{a^2+2ad+d^2}{a}-a+d=30</cmath>
+<cmath>\frac{(a+2d)a+d^2}{a}-a+d=30</cmath>
+<cmath>a+2d+\frac{d^2}{a}-a+d=30</cmath>
+Combining like terms we get
+<cmath>3d+\frac{d^2}{a}=30</cmath>
+And looking at the values we know that since the second term must be positive (since both a and d are positive), <math>d</math> must be a maximum of 9, which offers 9 possible values (since <math>d</math> must be a positive integer.) We can resort to guesswork by this time, and thus receive the result of of <math>d=9</math> at which <math>a=27</math>. The solution is thus <math>\boxed{129}.</math>
+* We could also lay down some further parameters for the value of <math>d</math>. as <math>a</math> must be greater than <math>d</math> (for the first term to be positive), we can substitite in a "borderline" value of a=d (which itself is just over the limit). This gives us:
+<cmath>3d+\frac{d^2}{d}=30</cmath>
+<cmath>3d+d=30</cmath>
+<cmath>4d=30</cmath>
+Thus, d must be greater than 7.5, which gives us only 2 values left (8 and 9). We can then plug in any of the two to see if an integer value is attainable. In the end, 8 doesn't work and 9 does, which gives the solution of <math>\boxed{129}.</math>
+* Note: This solution is similar to Solution 2
+-PMaldonado13
 == 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

Page

Toolbox

Search