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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
3 years, 8 months 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.

\(abc = 2(1 + \omega)(1 + \omega^2) = \fbox{2}\) Jatin Yadav · 3 years, 8 months ago

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

Edit: Pi Han Goh is actually the winner. Cody Johnson · 3 years, 8 months ago

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@Jatin Yadav The same process is followed be me! Shubham Kumar · 3 years, 8 months ago

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@Jatin Yadav Where did the omega come from? Bob Krueger · 3 years, 8 months ago

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@Bob Krueger \(\omega\) is usually used to denote \(e^{\pi i /3} = \text{cis} \dfrac{\pi}{3}\) (although he should have specified). Michael Tang · 3 years, 8 months ago

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@Michael Tang Michael, do you mean \(\omega=e^{2\pi i/3}\)? Kevin Chang · 3 years, 8 months ago

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@Michael Tang Thanks. Now it makes sense. Bob Krueger · 3 years, 8 months ago

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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 · 3 years, 8 months ago

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@Calvin Lin Actually, I believe Cody is asking for integer answers. Ahaan Rungta · 3 years, 8 months ago

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@Calvin Lin 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 · 3 years, 8 months ago

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@Cody Johnson 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 · 3 years, 8 months ago

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@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 · 3 years, 8 months ago

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@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 · 3 years, 8 months ago

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@Cody Johnson Yes, but that's not a good problem-solving tactic neither is it good problem-writing practice to do that. Ahaan Rungta · 3 years, 8 months ago

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yes 2 is the answer Anirudha Nayak · 3 years, 4 months ago

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\(3N = 2^{2013} -2\), 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 · 3 years, 8 months ago

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@George G \(3N\neq2^{2013}-2\) Cody Johnson · 3 years, 8 months ago

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@Cody Johnson 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 · 3 years, 8 months ago

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@Pi Han Goh Whoops, my bad. I guess it does. Cody Johnson · 3 years, 8 months ago

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@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 · 3 years, 8 months ago

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@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 · 3 years, 8 months ago

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@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 · 3 years, 8 months ago

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@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 · 3 years, 8 months ago

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@Jatin Yadav Thanks Jatin! :) Pranav Arora · 3 years, 8 months ago

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1 Sunitha Bhadragiri · 3 years, 8 months ago

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abc = 1 ? Joey Dandan · 3 years, 8 months ago

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