CMC - Problem 2

Problem 2. (5 points) Let

N=(20130)+(20133)+(20136)++(20132013).N=\binom{2013}{0}+\binom{2013}{3}+\binom{2013}{6}+\dots+\binom{2013}{2013}.

If 3N=a2013+b2013+c20133N=a^{2013}+b^{2013}+c^{2013} where a+b+c=3a+b+c=3, find the value of abc.abc.

Note by Cody Johnson
5 years, 11 months ago

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24 votes

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I did in this way:

(1+x)2013=(20130)+(20131)x+(20132)x2+(20132012)x2012+(20132013)x2013(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+ωx)2013+(1+ω2x)2013,(1 + x)^{2013} + (1 + \omega x)^{2013} + (1+ \omega^2 x)^{2013}, coeff. of x3k+1x^{3k + 1} and x3k+2x^{3k + 2} becomes 0 , and coeff. of x3kx^{3k} triples

Now, Replace x = 1,

(1+1)2013+(1+ω)2013+(1+ω2)2013(1 + 1)^{2013} + (1 + \omega )^{2013} + (1+ \omega^2 )^{2013}

= 3((20130)+(20133)+(20136)++(20132013))3\bigg({2013 \choose 0} + {2013 \choose 3} + {2013 \choose 6} + \dots + {2013 \choose 2013}\bigg)

= 3N3N

Hence, 3N=(1+1)2013+(1+ω)2013+(1+ω2)20133N = (1 + 1)^{2013} + (1 + \omega)^{2013} + ( 1+ \omega^2)^{2013}

Hence , a=2,b=1+ω,c=1+ω2a = 2, b = 1+\omega, c = 1 + \omega^2, a+b+c=3+(1+ω+ω2)=3a + b+ c = 3 + (1 + \omega + \omega^2) = 3, as required.

abc=2(1+ω)(1+ω2)=2abc = 2(1 + \omega)(1 + \omega^2) = \fbox{2}

jatin yadav - 5 years, 11 months ago

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

Edit: Pi Han Goh is actually the winner.

Cody Johnson - 5 years, 11 months ago

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The same process is followed be me!

SHUBHAM KUMAR - 5 years, 11 months ago

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Where did the omega come from?

Bob Krueger - 5 years, 11 months ago

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ω\omega is usually used to denote eπi/3=cisπ3e^{\pi i /3} = \text{cis} \dfrac{\pi}{3} (although he should have specified).

Michael Tang - 5 years, 11 months ago

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@Michael Tang Michael, do you mean ω=e2πi/3\omega=e^{2\pi i/3}?

Kevin Chang - 5 years, 11 months ago

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@Michael Tang Thanks. Now it makes sense.

Bob Krueger - 5 years, 11 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 - 5 years, 11 months ago

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Actually, I believe Cody is asking for integer answers.

Ahaan Rungta - 5 years, 11 months ago

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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 2\boxed{2} is the only number that satisfies the problem condition."

Note that I've computed that 22 is the only integer answer for this problem.

Cody Johnson - 5 years, 11 months ago

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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 abc.

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

Calvin Lin Staff - 5 years, 11 months ago

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@Calvin Lin So would it be better to say, "find the only possible integer value of abcabc?" Or maybe I should've asked for some completely different thing, like to find N(mod2011)N\pmod{2011}?

Cody Johnson - 5 years, 11 months ago

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@Cody Johnson I think it would have been better to ask for a closed form expression for NN, and 2201323 \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 abcabc", 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,ca, b, c in your question, but this need not be the only solution. It is possible for abc=0abc=0 , by setting c=0c=0 and a2013+(3a)2013=220132 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] [0, 1.5] (real interval). However, this strays away from the original intention of your question.

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

Calvin Lin Staff - 5 years, 11 months ago

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

Ahaan Rungta - 5 years, 11 months ago

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yes 2 is the answer

Anirudha Nayak - 5 years, 8 months ago

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3N=2201323N = 2^{2013} -2, one choice for a,b,ca,b,c are 2,x,x22, -x, -x^2 where x2+x+1=0x^2+x+1=0. So abc=2abc=2 in this case. But ..

George G - 5 years, 11 months ago

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3N2201323N\neq2^{2013}-2

Cody Johnson - 5 years, 11 months ago

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Actually, 3N=2201323N = 2^{2013} - 2 , I have proved that in my deleted comment. I've removed my comment because I couldn't prove that abc=2abc=2 only like what Jatin did.

Pi Han Goh - 5 years, 11 months ago

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@Pi Han Goh You are right, 3N=2201323N = 2^{2013} - 2

Actually, 3N=22013+(1+ω)2013+(1+ω2)20133N = 2^{2013} + (1 + \omega)^{2013} + (1 + \omega^2)^{2013}

= 22013+2Re(1+ω)2013=22013+2Re(eiπ3×2013) 2^{2013} + 2Re(1 + \omega)^{2013} = 2^{2013} + 2 Re(e^{i\frac{\pi}{3} \times 2013})

= 2201322^{2013} - 2

How did you prove it?

jatin yadav - 5 years, 11 months ago

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@Jatin Yadav I first consider N0=(00),N1=(30)+(33),N2=(60)+(63)+(66),,N671=(20130)+(20133)+(20132013) 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 3N0,3N1,3N2,3N33N_0, 3N_1, 3N_2, 3N_3 are very close to the powers of 88, so I made the conjecture 3Nj=8j+2(1)j3 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 - 5 years, 11 months ago

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@Jatin Yadav Hi Jatin!

How do you get (1+ω)2013+(1+ω2)2013=2Re(1+ω)2013 (1+\omega)^{2013}+(1+\omega^2)^{2013}=2Re(1+\omega)^{2013} ?

I did it this way, since 1+ω+ω2=01+\omega+\omega^2=0, hence we have 1+ω=ω21+\omega=-\omega^2 and 1+ω2=ω1+\omega^2=-\omega.

Therefore, (1+ω)2013+(1+ω2)2013=2(1+\omega)^{2013}+(1+\omega^2)^{2013}=-2.

Many thanks!

Pranav Arora - 5 years, 11 months ago

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@Pranav Arora Hi, (1+ω)2013(1+\omega)^{2013} is conjugate of (1+ω2)2013(1 + \omega^2)^{2013},

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

jatin yadav - 5 years, 11 months ago

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@Jatin Yadav Thanks Jatin! :)

Pranav Arora - 5 years, 11 months ago

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

Cody Johnson - 5 years, 11 months ago

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1

sunitha bhadragiri - 5 years, 11 months ago

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abc = 1 ?

Joey Dandan - 5 years, 11 months ago

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