# CMC - Problem 2

Problem 2. (5 points) Let

$N=\binom{2013}{0}+\binom{2013}{3}+\binom{2013}{6}+\dots+\binom{2013}{2013}.$

If $$3N=a^{2013}+b^{2013}+c^{2013}$$ where $$a+b+c=3$$, find the value of $$abc.$$

Note by Cody Johnson
5 years, 1 month ago

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I 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.

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

- 5 years, 1 month ago

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

Edit: Pi Han Goh is actually the winner.

- 5 years, 1 month ago

The same process is followed be me!

- 5 years, 1 month ago

Where did the omega come from?

- 5 years, 1 month ago

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

- 5 years, 1 month ago

Michael, do you mean $$\omega=e^{2\pi i/3}$$?

- 5 years, 1 month ago

Thanks. Now it makes sense.

- 5 years, 1 month ago

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.

Staff - 5 years, 1 month ago

- 5 years, 1 month ago

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.

- 5 years, 1 month ago

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.

Staff - 5 years, 1 month ago

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 $$N\pmod{2011}$$?

- 5 years, 1 month ago

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.

Staff - 5 years, 1 month ago

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

- 5 years, 1 month ago

- 4 years, 10 months ago

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

- 5 years, 1 month ago

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

- 5 years, 1 month ago

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.

- 5 years, 1 month ago

You are right, $$3N = 2^{2013} - 2$$

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?

- 5 years, 1 month ago

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

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.

Great job by the way!

- 5 years, 1 month ago

Hi Jatin!

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!

- 5 years, 1 month ago

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

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

- 5 years, 1 month ago

Thanks Jatin! :)

- 5 years, 1 month ago

Whoops, my bad. I guess it does.

- 5 years, 1 month ago

1

- 5 years, 1 month ago

abc = 1 ?

- 5 years, 1 month ago