Considering that you're using a discussion instead of requiring an integer answer, you should adjust your question accordingly.

The way your question is currently phrased, there are (likely to be) numerous answers. Think of a (complex) plane cutting a sucked in ball. We know that there are 6 points of intersection, and this plane is unlikely to be tangential at each of these points.

Tell me this - if an integer answer is required by the problem, doesn't that force the user to find the integer answer only? A line in the solution would be like, "And, since we're guaranteed an integer answer, then $\boxed{2}$ is the only number that satisfies the problem condition."

Note that I've computed that $2$ is the only integer answer for this problem.

The problem should stand along by itself, without requiring the reader to know that "only integer answers are allowed". If this is a necessary assumption, it should be clearly stated in the question that you are only looking for integer answers. The question, as stated, could have several answers in a neighborhood of 2. It could be interpreted as finding the range of possible values of $abc$.

Note that Jatin's solution only shows that $abc=2$ is a possible answer, you should explain why no other integer answers are possible.

@Calvin Lin
–
So would it be better to say, "find the only possible integer value of $abc$?" Or maybe I should've asked for some completely different thing, like to find $N\pmod{2011}$?

@Cody Johnson
–
I think it would have been better to ask for a closed form expression for $N$, and $\frac{2^{2013} -2 } {3}$ would have been a good answer. There isn't a strong reason for you to stick to "integer answers from 0 to 999". We do that simply because it is the simplest to explain, without having to deal with various other explanations / considerations.

By asking for "the only possible integer value of $abc$", you then require a proof that it is the only possible integer value, which need not necessarily be the case. Yes, there is a natural choice of $a, b, c$ in your question, but this need not be the only solution. It is possible for $abc=0$, by setting $c=0$ and $a^{2013} + (3-a)^{2013} = 2^{2013} -2$, and showing that a solution must exist by applying the intermediate value theorem on $[0, 1.5]$ (real interval). However, this strays away from the original intention of your question.

Note: I do not know if $abc=1$ is possible. It most likely is, but I can't think of an immediate argument for it.

Actually, $3N = 2^{2013} - 2$, I have proved that in my deleted comment. I've removed my comment because I couldn't prove that $abc=2$ only like what Jatin did.

And I find that $3N_0, 3N_1, 3N_2, 3N_3$ are very close to the powers of $8$, so I made the conjecture $3 N_j = 8^j + 2(-1)^j$ and I proved it by induction with the help of Pascal's identities, it was a little tedious.

$</code> ... <code>$</code>...<code>."> Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in $</span> ... <span>$ or $</span> ... <span>$ to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestI did in this way:

$(1 + x)^{2013} = {2013 \choose 0} + {2013 \choose 1}x + {2013 \choose 2}x^2+ \dots {2013 \choose 2012}x^{2012} + {2013 \choose 2013}x^{2013}$

Hence , in $(1 + x)^{2013} + (1 + \omega x)^{2013} + (1+ \omega^2 x)^{2013},$ coeff. of $x^{3k + 1}$ and $x^{3k + 2}$ becomes 0 , and coeff. of $x^{3k}$ triples

Now, Replace x = 1,

$(1 + 1)^{2013} + (1 + \omega )^{2013} + (1+ \omega^2 )^{2013}$

= $3\bigg({2013 \choose 0} + {2013 \choose 3} + {2013 \choose 6} + \dots + {2013 \choose 2013}\bigg)$

= $3N$

Hence, $3N = (1 + 1)^{2013} + (1 + \omega)^{2013} + ( 1+ \omega^2)^{2013}$

Hence , $a = 2, b = 1+\omega, c = 1 + \omega^2$, $a + b+ c = 3 + (1 + \omega + \omega^2) = 3$, as required.

$abc = 2(1 + \omega)(1 + \omega^2) = \fbox{2}$

Log in to reply

Exactly how I did it. Good work, you get the 5 points!

Edit: Pi Han Goh is actually the winner.

Log in to reply

The same process is followed be me!

Log in to reply

Where did the omega come from?

Log in to reply

$\omega$ is usually used to denote $e^{\pi i /3} = \text{cis} \dfrac{\pi}{3}$ (although he should have specified).

Log in to reply

$\omega=e^{2\pi i/3}$?

Michael, do you meanLog in to reply

Log in to reply

Considering that you're using a discussion instead of requiring an integer answer, you should adjust your question accordingly.

The way your question is currently phrased, there are (likely to be) numerous answers. Think of a (complex) plane cutting a sucked in ball. We know that there are 6 points of intersection, and this plane is unlikely to be tangential at each of these points.

Log in to reply

Actually, I believe Cody is asking for integer answers.

Log in to reply

Tell me this - if an integer answer is required by the problem, doesn't that force the user to find the integer answer only? A line in the solution would be like, "And, since we're guaranteed an integer answer, then $\boxed{2}$ is the only number that satisfies the problem condition."

Note that I've computed that $2$ is the only integer answer for this problem.

Log in to reply

The problem should stand along by itself, without requiring the reader to know that "only integer answers are allowed". If this is a necessary assumption, it should be clearly stated in the question that you are only looking for integer answers. The question, as stated, could have several answers in a neighborhood of 2. It could be interpreted as finding the range of possible values of $abc$.

Note that Jatin's solution only shows that $abc=2$ is a possible answer, you should explain why no other integer answers are possible.

Log in to reply

$abc$?" Or maybe I should've asked for some completely different thing, like to find $N\pmod{2011}$?

So would it be better to say, "find the only possible integer value ofLog in to reply

$N$, and $\frac{2^{2013} -2 } {3}$ would have been a good answer. There isn't a strong reason for you to stick to "integer answers from 0 to 999". We do that simply because it is the simplest to explain, without having to deal with various other explanations / considerations.

I think it would have been better to ask for a closed form expression forBy asking for "the only possible integer value of $abc$", you then require a proof that it is the only possible integer value, which need not necessarily be the case. Yes, there is a natural choice of $a, b, c$ in your question, but this need not be the only solution. It is possible for $abc=0$, by setting $c=0$ and $a^{2013} + (3-a)^{2013} = 2^{2013} -2$, and showing that a solution must exist by applying the intermediate value theorem on $[0, 1.5]$ (real interval). However, this strays away from the original intention of your question.

Note: I do not know if $abc=1$ is possible. It most likely is, but I can't think of an immediate argument for it.

Log in to reply

Yes, but that's not a good problem-solving tactic neither is it good problem-writing practice to do that.

Log in to reply

yes 2 is the answer

Log in to reply

$3N = 2^{2013} -2$, one choice for $a,b,c$ are $2, -x, -x^2$ where $x^2+x+1=0$. So $abc=2$ in this case. But ..

Log in to reply

$3N\neq2^{2013}-2$

Log in to reply

Actually, $3N = 2^{2013} - 2$, I have proved that in my deleted comment. I've removed my comment because I couldn't prove that $abc=2$ only like what Jatin did.

Log in to reply

$3N = 2^{2013} - 2$

You are right,Actually, $3N = 2^{2013} + (1 + \omega)^{2013} + (1 + \omega^2)^{2013}$

= $2^{2013} + 2Re(1 + \omega)^{2013} = 2^{2013} + 2 Re(e^{i\frac{\pi}{3} \times 2013})$

= $2^{2013} - 2$

How did you prove it?

Log in to reply

$N_0 = {0 \choose 0}, N_1 = {3 \choose 0} + { 3 \choose 3 }, N_2 = {6 \choose 0 } + {6 \choose 3 } + { 6 \choose 6} , \ldots , N_{671} = {2013 \choose 0 } + {2013 \choose 3 } + \ldots {2013 \choose 2013}$

I first considerAnd I find that $3N_0, 3N_1, 3N_2, 3N_3$ are very close to the powers of $8$, so I made the conjecture $3 N_j = 8^j + 2(-1)^j$ and I proved it by induction with the help of Pascal's identities, it was a little tedious.

Great job by the way!

Log in to reply

How do you get $(1+\omega)^{2013}+(1+\omega^2)^{2013}=2Re(1+\omega)^{2013}$?

I did it this way, since $1+\omega+\omega^2=0$, hence we have $1+\omega=-\omega^2$ and $1+\omega^2=-\omega$.

Therefore, $(1+\omega)^{2013}+(1+\omega^2)^{2013}=-2$.

Many thanks!

Log in to reply

$(1+\omega)^{2013}$ is conjugate of $(1 + \omega^2)^{2013}$,

Hi,and $z + \bar{z} = 2Re(z)$, well known identity in complex nos.

Log in to reply

Log in to reply

Log in to reply

1

Log in to reply

abc = 1 ?

Log in to reply