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

Revision as of 16:06, 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 $c = a+2 = 925 \cdot 27 - 2)+2 = 25 \cdot 27 = \boxed{675}.$

@@ Line 35: / Line 35: @@
-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 = 925 \cdot 27 - 2)+2 = 25\ cdot 27 = \boxed{675}.</math>
+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 = 925 \cdot 27 - 2)+2 = 25 \cdot 27 = \boxed{675}.</math>
 == See also ==
 {{AIME box|year=1989|num-b=3|num-a=5}}
 {{MAA Notice}}

1989 AIME (Problems • Answer Key • Resources)
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