The product of three consecutive odd integers is -6783. How do you write and solve an equation to find the numbers?

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The product of three consecutive odd integers is -6783. How do you write and solve an equation to find the numbers?

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Answer 1 · 2016-07-18T05:45:51

$- 21, - 19, - 17$ $-21,-19,-17$

Explanation:

This problem can be solved by using some pretty nifty algebra.

Effectively the problem is $a \cdot b \cdot c = - 6783$ $a*b*c=-6783$ solve for $a, b,$ $a, b,$ and $c$ $c$ . However we can rewrite $b$ $b$ and $c$ $c$ in terms of $a$ $a$ . We do this by thinking what consecutive odd numbers are.

For instance, $1, 3,$ $1, 3,$ and $5$ $5$ are 3 consecutive odd numbers, the difference between $1$ $1$ and $3$ $3$ is $2$ $2$ , and the difference between $5$ $5$ and $1$ $1$ is $4$ $4$ . So if we write it in terms of $1$ $1$ , the numbers would be $1, 1 + 2,$ $1, 1+2,$ and $1 + 4$ $1+4$ .

Now lets bring it back to variables and put it in terms of $a$ $a$ . $b$ $b$ would just equal $a + 2$ $a+2$ being the next odd number, and the number after that, $c$ $c$ , would just equal $a + 4$ $a+4$ . So now lets plug this into $a \cdot b \cdot c = - 6783$ $a*b*c=-6783$ and let's solve.

$(a) (a + 2) (a + 4) = - 6783$ $(a)(a+2)(a+4)=-6783$
$(a^{2} + 2 a) (a + 4) = - 6783$ $(a^2+2a)(a+4)=-6783$
$a^{3} + 4 a^{2} + 2 a^{2} + 8 a = - 6783$ $a^3+4a^2+2a^2+8a=-6783$
$a^{3} + 6 a^{2} + 8 a + 6783 = 0$ $a^3+6a^2+8a+6783=0$

Now from here I'm going to graph looking for possible values for $a$ $a$ . The jist of this is to graph $a^{3} + 6 a^{2} + 8 a + 6783$ $a^3+6a^2+8a+6783$ and find where the equation is equal to $0$ $0$ .

graph{x^3+6x^2+8x+6783 [-207.8, 207.7, -108.3, 108.3]}

As you can see it's a pretty big graph so I'm only going to show the meaningful part, the intersection. Here we can see that the graph intersects at $a = - 21$ $a = -21$ , you can click on the graph yourself to find it.

So if -21 is our starting number, our following numbers will be -19 and -17. Let's test shall we?

$- 21 \cdot - 19 = 399$ $-21*-19=399$
$399 \cdot - 17 = - 6783$ $399*-17=-6783$

Excellent!

Now upon research to ensure that I was doing this a good way, I actually found a trick on this website was a short little trick someone found. If you take the cube root of the product and round the number to the nearest whole integer, you will find the middle odd number. The cube root of $- 6783$ $-6783$ is $- 18.929563765$ $-18.929563765$ which rounds to $- 19$ $-19$ . Hey that's the middle number that we found right?

Now about that trick, I'm not quite sure how reliable it is under all circumstances but if you have a calculator (which with this algebra I hope you do), maybe use it to check.

Answer 2 · 2016-07-18T06:06:47

If you do not have to show specific algebraic work (and especially if you can use a calculator (think SAT)), this particular problem lends well to a sneaky shortcut.

Explanation:

Since there are three unknown values which are consecutive odds and thus all very close to each other...

What is the cube-root of $6783$ $6783$ ? (Use calculator.) Approximately $18.92956...$ The nearest odd number to that is $19$ , and its nearest odd neighbors are $17$ and $21$ . So, try those three and see what happens. $17*19*21=6783$ . Nice.

Oh, but we wanted $-6783$ , so make it $-17$ , $-19$ , and $-21$ . Done.

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The product of three consecutive odd integers is -6783. How do you write and solve an equation to find the numbers?

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