Binomial coefficient: Search for multiples of 13

Question

Question: The coefficients of how many terms in the expansion of (1+x)²⁰¹⁸ are multiples of 13?

So, we've to investigate the powers of 13 in ${2018 \choose r}$, where 0 ≤ r ≤ 2018

I tried using the following: $$s_p(N!) = \left \lfloor \frac{N}{p} \right \rfloor + \left \lfloor \frac{N}{p^2} \right \rfloor + \left \lfloor \frac{N}{p^3} \right \rfloor + \cdots$$

where, $s_p(N!)$ denotes the highest exponent of prime p in n!, where n is a natural number.

${2018 \choose r}$, may be written as $\frac{2018!}{r!(2018-r)!}$. Now, I tried to figure out the exponent of 13 in numerator for several values of r, but couldn't find a pattern that would lead me to the desired answer.

Could someone please give a detailed solution to this problem, and also explain how to approach such problems? Is there any generalisation for the multiples of a prime number p in ${n \choose r}$?

P.S. The answer, to the best of my recollection, is 1395.

It can generally be solved using Luca's theorem or the Kummer's theorem. You can search on Google for more information about them — Rohan Shinde, Feb 21 '18 at 05:22
Great. Thanks a lot. Can we not solve it using the idea I proposed, though? — stoic-santiago, Feb 21 '18 at 05:24
I do have another method . I will just post it as hint. You try to figure out the underlying answer. OK? — Rohan Shinde, Feb 21 '18 at 05:28
Yeah, sure. Post it as an answer though. I'll try to figure it out. — stoic-santiago, Feb 21 '18 at 05:32

Rohan Shinde · Accepted Answer · 2018-02-21T07:17:19.160

1

Hint:

$$E_{13}(2018!)=\left \lfloor \frac {2018}{13}\right \rfloor+\left \lfloor \frac {2018}{13^2}\right \rfloor+\left \lfloor \frac {2018}{13^3}\right \rfloor \ldots $$ $$=155+11+0$$ $$=166$$

$$\Rightarrow E_{13}(r!) +E_{13}((2018-r)!)\le 165$$ $$\Rightarrow r=13\lambda +k, k=4,5,6,7,8,9,10,11,12$$

For $r=13\lambda+ 4$ we have following $155$ numbers $$r=4,17,30,....., 2006$$

And continue similarly for other values of $k$

$$\Rightarrow Total = 155*9=1395$$

edited Feb 21 '18 at 07:17

answered Feb 21 '18 at 05:34

Rohan Shinde

9,737

@TobiasKildetoft Sorry for the typo. And thanks , it is fixed now. – Rohan Shinde Feb 21 '18 at 07:18
@schrodinger_16 its not just of the form 13m+4. it is of the the form 13m+k where k varies from 4 to 12. we get this condition when we solve the so formed inequality i gave. the general solution of the inequality was $$r\equiv 4,5,6,7,8,9,10,11,12 \pmod {13}$$ – Rohan Shinde Feb 21 '18 at 08:13
Great. Thanks, got it. This is the kind of the solution I was looking for! – stoic-santiago Feb 21 '18 at 08:20

score 0 · Answer 2 · answered Feb 21 '18 at 08:19

According to Kummer's Theorem, as shown in this answer,

The number of factors of $p$ in $\binom{n}{k}$ is the number of carries in the addition $k+(n-k)=n$ when performed in base-$p$.

This means that the binomial coefficients $\binom{2018}{k}$ that are not divisible by $13$ are those where $k+(BC3_{13}-k)=BC3_{13}$ has no carries. That would be those with digits $0$-$B$ in the $13^2$ digit, $0$-$C$ in the $13^1$ digit, and $0$-$3$ in the $13^0$ digit. That is, there are $12\cdot13\cdot4=624$ coefficients which do not have a factor of $13$.

Thus, there are $2019-12\cdot13\cdot4=1395$ coefficients that have a factor of $13$.

Binomial coefficient: Search for multiples of 13

2 Answers2