# 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
6 years, 7 months ago

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

• Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
• Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
• Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears 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} $\boxed{123}$

Sort by:

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.

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

- 6 years, 7 months ago

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

Edit: Pi Han Goh is actually the winner.

- 6 years, 7 months ago

The same process is followed be me!

- 6 years, 7 months ago

Where did the omega come from?

- 6 years, 7 months ago

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

- 6 years, 7 months ago

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

- 6 years, 7 months ago

Thanks. Now it makes sense.

- 6 years, 7 months 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 - 6 years, 7 months ago

- 6 years, 7 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.

- 6 years, 7 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.

Staff - 6 years, 7 months 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 $N\pmod{2011}$?

- 6 years, 7 months 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 - 6 years, 7 months ago

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

- 6 years, 7 months ago

- 6 years, 3 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 ..

- 6 years, 7 months ago

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

- 6 years, 7 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.

- 6 years, 7 months 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?

- 6 years, 7 months 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!

- 6 years, 7 months 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!

- 6 years, 7 months ago

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

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

- 6 years, 7 months ago

Thanks Jatin! :)

- 6 years, 7 months ago

Whoops, my bad. I guess it does.

- 6 years, 7 months ago

1

- 6 years, 7 months ago

abc = 1 ?

- 6 years, 7 months ago