Largest prime no. by which $\left(\binom{60}{30}-1\right)$ is divisible, is

Question

[1] :: Largest prime no. by which $\displaystyle \left(\binom{60}{30}-1\right)$ is divisible, is

[2]:: Calculation of remainder when $\displaystyle \binom{2013}{101}$ is divided by $101$

$\bf{My\; Try}$ for (1):: Using $\displaystyle \binom{2n}{n} = \frac{2^n\cdot (1\cdot 3 \cdot 5 \cdot \ldots \cdot (2n-3)\cdot (2n-1))}{n!}$

So Put $n=30$, we get $\displaystyle \binom{60}{30} = \frac{2^{30}\cdot (1\cdot 3\cdot 5 \cdot \ldots \cdot 27\cdot 29)}{30!}$

Now I did not understand how can i solve it

Help Required

Thanks

Using Wolfram Alpha I get that ${60\choose 30}-1=3\times 61\times 646254544070281$. I have no idea how one might get this without serious computation. — vadim123, Oct 22 '13 at 03:39
Knowing the prime divisors of $N$ does little for getting the prime divisors of $N-1$ (except for discarding a few primes). — Carlos Eugenio Thompson Pinzón, Oct 22 '13 at 03:43
b.t.w. $$\binom{60}{30}=\frac{2^{30}\cdot(1\cdot3\cdot5\cdots57\cdot59)}{30!}$$. — Carlos Eugenio Thompson Pinzón, Oct 22 '13 at 03:55
It is easy to see that if $n$ is even and $2n+1=p$ is a prime, then ${2n\choose n}-1$ is divisible by $2n+1$. That explains the factor $61$ observed by Vadim. — Jyrki Lahtonen, Oct 22 '13 at 05:59

score 2 · Answer 1 · edited Apr 13 '17 at 12:21

2

The first question looks prohibitively difficult. Because $101$ is a prime the second question OTOH is a straightforward application of Lucas' correspondence. See my earlier answer for an on-site proof. Here the numbers in base $101$ are $$2013=19\cdot101+94,\quad 101=1\cdot101+0,$$ so Lucas tells us that $$ {2013\choose 101}\equiv{19\choose1}{94\choose 0}=19\cdot1=19\pmod{101}. $$

edited Apr 13 '17 at 12:21

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answered Oct 22 '13 at 05:45

Jyrki Lahtonen

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Largest prime no. by which $\left(\binom{60}{30}-1\right)$ is divisible, is

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