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

Revision as of 11:40, 3 March 2024

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 $c = a+2 = (25 \cdot 27 - 2)+2 = 25 \cdot 27 = \boxed{675}.$

Solution 4

(This is literally a combination of 1 and 3)

Since $a$ , $b$ , $c$ , $d$ , and $e$ are consecutive, $a = c-2$ , $b = c-1$ , $c=c$ , $d = c+1$ , and $e = c+2$ .

Because $b+c+d = 3c$ is a perfect square, and $a+b+c+d+e = 5c$ is a perfect cube, we can express $c$ as $c = 3^{n} \cdot 5^{k}$ .

Now, by the problem's given information,

$k \equiv 0 \text{(mod 2)}$

$n \equiv 0 \text{(mod 3)}$

and because ALL exponents have to be cubes/squares,

$k \equiv 2 \text{(mod 3)}$

$n \equiv 1 \text{(mod 2)}$

Therefore,

$k = 2$

$n = 3$

$c = 3^3 \cdot 5^2 = \boxed{675}$ .

@@ 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 = (25 \cdot 27 - 2)+2 = 25 \cdot 27 = \boxed{675}.</math>
+== Solution 4 ==
+(This is literally a combination of 1 and 3)
+Since <math>a</math>, <math>b</math>, <math>c</math>, <math>d</math>, and <math>e</math> are consecutive, <math>a = c-2</math>, <math>b = c-1</math>, <math>c=c</math>, <math>d = c+1</math>, and <math>e = c+2</math>.
+Because <math>b+c+d = 3c</math> is a perfect square, and <math>a+b+c+d+e = 5c</math> is a perfect cube, we can express <math>c</math> as <math>c = 3^{n} \cdot 5^{k}</math>.
+Now, by the problem's given information,
+<math>k \equiv 0   \text{(mod 2)}</math>
+<math>n \equiv 0   \text{(mod 3)}</math>
+and because ALL exponents have to be cubes/squares,
+<math>k \equiv 2   \text{(mod 3)}</math>
+<math>n \equiv 1   \text{(mod 2)}</math>
+Therefore,
+<math>k = 2</math>
+<math>n = 3</math>
+<math>c = 3^3 \cdot 5^2 = \boxed{675}</math>.
 == See also ==
 {{AIME box|year=1989|num-b=3|num-a=5}}
 {{MAA Notice}}

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
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