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 $abc.$

#CMC #Competitions #MathProblem #Math

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Comments

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.

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

jatin yadav - 7 years, 6 months ago

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

Edit: Pi Han Goh is actually the winner.

Cody Johnson - 7 years, 6 months ago

The same process is followed be me!

SHUBHAM KUMAR - 7 years, 6 months ago

Where did the omega come from?

Bob Krueger - 7 years, 6 months ago

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

Michael Tang - 7 years, 6 months ago

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

Kevin Chang - 7 years, 6 months ago

@Michael Tang – Thanks. Now it makes sense.

Bob Krueger - 7 years, 6 months ago

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.

Calvin Lin Staff - 7 years, 6 months ago

Actually, I believe Cody is asking for integer answers.

Ahaan Rungta - 7 years, 6 months 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.

Cody Johnson - 7 years, 6 months 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.

Calvin Lin Staff - 7 years, 6 months ago

@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 - 7 years, 6 months ago

@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.

Calvin Lin Staff - 7 years, 6 months ago

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

Ahaan Rungta - 7 years, 6 months ago

yes 2 is the answer

Anirudha Nayak - 7 years, 2 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 ..

George G - 7 years, 6 months ago

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

Cody Johnson - 7 years, 6 months 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.

Pi Han Goh - 7 years, 6 months ago

@Pi Han Goh – 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?

jatin yadav - 7 years, 6 months ago

@Jatin Yadav – 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!

Pi Han Goh - 7 years, 6 months ago

@Jatin Yadav – 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!

Pranav Arora - 7 years, 6 months ago

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

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

jatin yadav - 7 years, 6 months ago

@Jatin Yadav – Thanks Jatin! :)

Pranav Arora - 7 years, 6 months ago

@Pi Han Goh – Whoops, my bad. I guess it does.

Cody Johnson - 7 years, 6 months ago

sunitha bhadragiri - 7 years, 6 months ago

abc = 1 ?

Joey Dandan - 7 years, 6 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$