Question #0d004

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Question #0d004

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Answer 1 · 2017-07-13T03:51:25

The question seems to be wrong.

Explanation:

$a + b + c = 0$ $a+b+c=0$ lets check the value of $a^{2017} + b^{2017} + c^{2017}$ $a^2017+ b^2017 +c^2017$ for 2 sets of solutions.
If a=b=c=0 then, $a^{2017} + b^{2017} + c^{2017}$ $a^2017+ b^2017 +c^2017$ = 0 but when $a = 1, b = 1, c = - 2$ $a=1,b=1,c=-2$ then $a^{2017} + b^{2017} + c^{2017}$ $a^2017+ b^2017 +c^2017$ = $1 + 1 + {(- 2)}^{2017}$ $1+1+(-2)^2017$ ≠ 0

Answer 2 · 2017-07-13T08:40:07

There are infinitely many answers.

Explanation:

If $a + b + c = 0$ $a + b + c = 0$ , then the addition of the variables has to add to zero.

Since you didn't say that $a^{2017} + b^{2017} + c^{2017} = 0$ $a^2017 + b^2017 + c^2017 = 0$ , I can't argue that each variable could be $0$ $0$ . However, I could say that

$a^{2017} + b^{2017} + c^{2017} = x$ $a^2017 + b^2017 + c^2017 = x$

which can also be written as

$a^{2017} + b^{2017} + c^{2017} - x = 0$ $a^2017 + b^2017 + c^2017 - x = 0$

which would give a lot more leeway in an answer, because then you're just working for a constant $x$ $x$ .

One answer (because there are now infinitely many) is if

$a = - 1$ $a = -1$
$b = - 1$ $b = -1$
$c = 2$ $c = 2$
$c^{2017} \approx 1.5 \cdot 10^{607}$ $c^2017 ~~ 1.5 * 10^607$ ^
$x \approx 1.5 \cdot 10^{607}$ $x ~~ 1.5 * 10^607$ ^^

then

$a + b + c = 0$ $a + b + c = 0$

$- 1 + (- 1) + 2 = 0$ $-1 + (-1) + 2 = 0$

From there, you have

${(- 1)}^{2017} + {(- 1)}^{2017} + 2^{2017} - (1.5 \cdot 10^{607})$ $(-1)^2017 + (-1)^2017 + 2^2017 - (1.5 * 10^607)$

The one problem here is that, of course, ${(- 1)}^{2017} = - 1$ $(-1)^2017 = -1$ . So, when using basic approximations with such large numbers, you'd still get $- 2$ $-2$ as an answer, unless of course you were exactly precise. Wolfram Alpha (see notes below) tells me that there are 608 decimal places in $2^{2017}$ $2^2017$ , so you'd be doing a lot of work for not a lot of payoff.

The point here, however, is that there is an answer to this problem--in fact, there's infinitely many--but it's much too complex for anyone to do by hand using basic algebra. There might be a way to do this with calculus and limits, though I will freely admit I'm not sure what that would be.

Notes:

^: See https://www.wolframalpha.com/input/?i=2%5E2017 for the scientific notation of $2^{2017}$ $2^2017$ .
^^: See https://www.wolframalpha.com/input/?i=2%5E2017+%2B+2 for the scientific notation of $2^{2017} + 2$ $2^2017 + 2$ . Hint: It's practically the same number at that point.

Edit: Of course, it just occurred to me that with the equation of $a^{2017} + b^{2017} + c^{2017} - x = 0$ $a^2017 + b^2017 + c^2017 - x = 0$ , you could hypothetically just say that each variable equals $0$ $0$ and any string of zeroes added/subtracted together will yield $0$ $0$ , which shortens this answer considerably.

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Question #0d004

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