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 "1989 AIME Problems/Problem 4"

m
(→‎Solution 3)
(7 intermediate revisions by 6 users not shown)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
−
If <math>a<b<c<d<e^{}_{}</math> are consecutive positive integers such that <math>b+c+d^{}_{}</math> is a perfect square and <math>a+b+c+d+e^{}_{}</math> is a perfect cube, what is the smallest possible value of <math>c^{}_{}</math>?
+
If <math>a<b<c<d<e</math> are [[consecutive]] [[positive]] [[integer]]s such that <math>b+c+d</math> is a [[perfect square]] and <math>a+b+c+d+e</math> is a [[perfect cube]], what is the smallest possible value of <math>c</math>?
  
 
== Solution ==
 
== Solution ==
−
{{solution}}
+
Since the middle term of an [[arithmetic progression]] with an odd number of terms is the average of the series, we know <math>b + c + d = 3c</math> and <math>a + b + c + d + e = 5c</math>. Thus, <math>c</math> must be in the form of <math>3 \cdot x^2</math> based upon the first part and in the form of <math>5^2 \cdot y^3</math> based upon the second part, with <math>x</math> and <math>y</math> denoting an [[integer]]s. <math>c</math> is minimized if it’s [[prime factorization]] contains only <math>3,5</math>, and since there is a cubed term in <math>5^2 \cdot y^3</math>, <math>3^3</math> must be a factor of <math>c</math>. <math>3^35^2 = \boxed{675}</math>, which works as the solution.
 +
 
 +
 
 +
==Solution 2==
 +
Let <math>b</math>, <math>c</math>, <math>d</math>, and <math>e</math> equal <math>a+1</math>, <math>a+2</math>, <math>a+3</math>, and <math>a+4</math>, respectively. Call the square and cube <math>k^2</math> and <math>m^3</math>, where both k and m are integers. Then:
 +
 
 +
<math>5a + 10 = m^3</math>
 +
 
 +
Now we know <math>m^3</math> is a multiple of 125 and <math>m</math> is a multiple of 5. The lower <math>m</math> is, the lower the value of <math>c</math> will be. Start from 5 and add 5 each time.
 +
 
 +
<math>m = 5</math> gives no solution for k
 +
 
 +
<math>m = 10</math> gives no solution for k
 +
 
 +
<math>m = 15</math> gives a solution for k.
 +
 
 +
 
 +
<math>10 + 5a = 15^3</math>
 +
 
 +
 
 +
<math>2 + a = 675</math>
 +
 
 +
 
 +
<math>c = \boxed{675}</math>
 +
 
 +
 
 +
-jackshi2006
 +
 
 +
 
 +
== Solution 3 ==
 +
 
 +
 
 +
Let the numbers be <math>a,a+1,a+2,a+3,a+4.</math> When then know <math>3a+6</math> is a perfect cube and <math>5a+10</math> is perfect cube. Since <math>5a+10</math> is divisible by <math>5</math> we know that <math>5a+10 = (5k)^3</math> since otherwise we get a contradiction. This means <math>a = 25k^3 - 2</math> in which plugging into the other expression we know <math>3(25k^3 - 2) + 6 = 75k^3</math> is a perfect square. We know <math>75 = 5^2 \cdot 3</math> so we let <math>k = 3</math> to obtain the perfect square. This means that <math>c = a+2 = (25 \cdot 27 - 2)+2 = 25 \cdot 27 = \boxed{675}.</math>
  
 
== See also ==
 
== See also ==
−
* [[1989 AIME Problems/Problem 5|Next Problem]]
+
{{AIME box|year=1989|num-b=3|num-a=5}}
−
* [[1989 AIME Problems/Problem 3|Previous Problem]]
+
{{MAA Notice}}
−
* [[1989 AIME Problems]]
 

Revision as of 16:13, 14 September 2020

Problem

If $a<b<c<d<e$ are consecutive positive integers such that $b+c+d$ is a perfect square and $a+b+c+d+e$ is a perfect cube, what is the smallest possible value of $c$?

Solution

Since the middle term of an arithmetic progression with an odd number of terms is the average of the series, we know $b + c + d = 3c$ and $a + b + c + d + e = 5c$. Thus, $c$ must be in the form of $3 \cdot x^2$ based upon the first part and in the form of $5^2 \cdot y^3$ based upon the second part, with $x$ and $y$ denoting an integers. $c$ is minimized if it’s prime factorization contains only $3,5$, and since there is a cubed term in $5^2 \cdot y^3$, $3^3$ must be a factor of $c$. $3^35^2 = \boxed{675}$, which works as the solution.


Solution 2

Let $b$, $c$, $d$, and $e$ equal $a+1$, $a+2$, $a+3$, and $a+4$, respectively. Call the square and cube $k^2$ and $m^3$, where both k and m are integers. Then:

$5a + 10 = m^3$

Now we know $m^3$ is a multiple of 125 and $m$ is a multiple of 5. The lower $m$ is, the lower the value of $c$ will be. Start from 5 and add 5 each time.

$m = 5$ gives no solution for k

$m = 10$ gives no solution for k

$m = 15$ gives a solution for k.


$10 + 5a = 15^3$


$2 + a = 675$


$c = \boxed{675}$


-jackshi2006


Solution 3

Let the numbers be $a,a+1,a+2,a+3,a+4.$ When then know $3a+6$ is a perfect cube and $5a+10$ is perfect cube. Since $5a+10$ is divisible by $5$ we know that $5a+10 = (5k)^3$ since otherwise we get a contradiction. This means $a = 25k^3 - 2$ in which plugging into the other expression we know $3(25k^3 - 2) + 6 = 75k^3$ is a perfect square. We know $75 = 5^2 \cdot 3$ so we let $k = 3$ to obtain the perfect square. This means that $c = a+2 = (25 \cdot 27 - 2)+2 = 25 \cdot 27 = \boxed{675}.$

See also

1989 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
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=1989_AIME_Problems/Problem_4&oldid=133651"