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Difference between revisions of "2003 AIME I Problems/Problem 8"
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Both <math>a</math> and <math>d</math> are positive integers, so <math>2d - 15</math> and <math>10 - d</math> must have the same sign. Try if they are both [[positive]] (notice if they are both [[negative]], then <math>d > 10</math> and <math>d < \frac{15}{2}</math>, which is a contradiction). Then, <math>d = 8, 9</math>. Directly substituting and testing shows that <math>d \neq 8</math>, but that if <math>d = 9</math> then <math>a = 18</math>. Alternatively, note that <math>3|2d</math> or <math>3|2d-15</math> implies that <math>3|d</math>, so only <math>9</math> may work. Hence, the four terms are <math>18,\ 27,\ 36,\ 48</math>, which indeed fits the given conditions. Their sum is <math>\boxed{129}</math>.
 
Both <math>a</math> and <math>d</math> are positive integers, so <math>2d - 15</math> and <math>10 - d</math> must have the same sign. Try if they are both [[positive]] (notice if they are both [[negative]], then <math>d > 10</math> and <math>d < \frac{15}{2}</math>, which is a contradiction). Then, <math>d = 8, 9</math>. Directly substituting and testing shows that <math>d \neq 8</math>, but that if <math>d = 9</math> then <math>a = 18</math>. Alternatively, note that <math>3|2d</math> or <math>3|2d-15</math> implies that <math>3|d</math>, so only <math>9</math> may work. Hence, the four terms are <math>18,\ 27,\ 36,\ 48</math>, which indeed fits the given conditions. Their sum is <math>\boxed{129}</math>.
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Postscript
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As another option, <math>3ad + 4d^2 = 30a + 30d</math> could be rewritten as follows:
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<math>d(3a + 4d) = 30(a + d)</math>
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<math>d(3a + 3d)+ d^2 = 30(a + d)</math>
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<math>3d(a + d)+ d^2 = 30(a + d)</math>
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<math>(3d - 30)(a + d)+ d^2 = 0</math>
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<math>3(d - 10)(a + d)+ d^2 = 0</math>
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This gives another way to prove <math>d<10</math>, and when rewritten one last time:
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<math>3(10 -d)(a + d) = d^2</math>
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shows that <math>d</math> must contain a factor of 3.
  
 
== See also ==
 
== See also ==

Revision as of 12:58, 7 July 2020

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.

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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