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Difference between revisions of "1989 AIME Problems/Problem 4"

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== 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.
 
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.
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==Solution 2==
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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:
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<math>5a + 10 = m^3</math>
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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.
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<math>m = 5</math> gives no solution for k
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<math>m = 10</math> gives no solution for k
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<math>m = 15</math> gives a solution for k.
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<math>10 + 5a = 15^3</math>
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<math>2 + a = 675</math>
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<math>c = \boxed{675}</math>
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 +
-jackshi2006
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=1989|num-b=3|num-a=5}}
 
{{AIME box|year=1989|num-b=3|num-a=5}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 12:59, 30 August 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

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

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