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- Feb 14, 2012

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- Feb 14, 2012

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- Feb 14, 2012

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$\displaystyle {2012\choose k}={4p\choose k}=\dfrac{(4p)!}{k!(4p-k)!}=\dfrac{4p}{k}\times {4p-1\choose k-1}$

Hence $\displaystyle {2012\choose k}=\dfrac{2012!}{k!(2012-k)!}$ is a multiple of $p$.

If $k$ is a multiple of $p$, there are only five cases to consider:

$k=0,\,p,\,2p,\,3p,\,4p$.

And we see that in all cases, the combination numbers

$\displaystyle {4p\choose 0}={4p\choose 4p}=1,\\{4p\choose p}={4p\choose 3p}=\dfrac{4(3p+1)(3p+2)\cdots(3p+(p-1))}{(p-1)!}\equiv 4 \pmod{p}$

and

$\displaystyle {4p\choose 2p}=\dfrac{6(2p+1)(2p+2)\cdots(2p+(p-1))[(2p+(p+1))(2p+(p+2))\cdots(2p+(2p-1))}{(p-1)![(p+1)(p+2)\cdots(2p-1)}\equiv 6 \pmod{p}$

are not multiple of $p$.

Now we denote the binary numeral of non-negative integer $n$ by

$\displaystyle n=(a_ra_{r-1}\cdots a_0)_2=\sum_{j=0}^r a_j2^j$ and $\displaystyle s(n)=\sum_{j=0}^r a_j$

where $a_j=0$ or 1 for $j=0,\,1,\,cdots,\,r$.

Then the power $m$ of factor $2^m$ in factorization of $n!$ can be expressed by

$\displaystyle \lfloor \dfrac{n}{2}\rfloor+\lfloor \dfrac{n}{4}\rfloor+\cdots +\lfloor \dfrac{n}{2^m}\rfloor+\cdots\\=(a_ra_{r-1}\cdots a_2a_1)_2+(a_ra_{r-1}\cdots a_3a_2)_2+\cdots+a_r\\=a_r\times (2^r-1)+a_{r-1}\times(2^{r-1}-1)+\cdots+a_1\times (2^1-1)+a_0\times (2^0-1)\\=n-s(n)$

If the combination number $\displaystyle {2012\choose k}=\dfrac{2012!}{k!(2012-k)!}=\dfrac{(k+m)!}{k!\times m!}$ is odd $(m=2012-k)$, then it means that the powers of 2 in factors of the numerator and the denominator of the fraction above are the same. Then

$k+m-s(k+m)=k-s(k)+m-s(m)$, or $s(k+m)=s(k)+s(m)$,

This means that the binary addition of $k+m=2012$ has no carrying.

Since $2012=(11111011100)_2$, it consists of eight 1's and three 0's. If there is no carrying in the addition of $k+m=(11111011100)_2$, then on the bit of 0 in $(11111011100)_2$, $k,\,m$ are 0, on the bit of 1, one is 1 and the other is 1, so there are two choices, $1=1+0$ and $1=0+1$. Thus, there are $2^8=256$ cases that the binary addition of two non-negative integers $k+m=2012$ has no carrying. That is, there are 256 combination numbers that are odd, and the remaining $2013-256=1757$ combination numbers are even.

If the combination number $\displaystyle {2012\choose k}=\dfrac{2012!}{k!(2012-k)!}=\dfrac{(k+m)!}{k!\times m!}$ is even but no a multiple of 4, then it means that the power of 2 in the numerator is greater than that in the denominator by 1. Thus,

$k+m-s(k+m)=k-s(k)+m-s(m)-1$, or $s(k+m)=s(k)+s(m)-1$.

That is, there is only one carrying in the binary addition of $k+m=2012$. The carrying happens at two bits as $01+01=10$.

By $k+m=2012=(11111011100)_2$, we see that the carrying can only happen at the fifth (from the highest bit to the lowest bit) and sixth bits or at the ninth and tenth bits. So there exist $2^7=128$ cases. That is, there are 256 combinations whose values are even numbers but not the multiples of 4.

Thus there are $2013-256-256=1501$ combinations whose values are multiples of 4. Now, we go back to consider the cases where $k$ is not the multiple of $p=503$.

We see that $\displaystyle {4p\choose 0}={4p\choose 4p}=1$ is not a multiple of 4. For $\displaystyle {4p\choose p}={4p\choose 3p}=\dfrac{(4p)!}{(2p)!(2p)!}$, the power of 2 is $s(2p)+s(2p)-s(4p)=s(p)=8$. Thus, there are three combination numbers which are multiples of 4 but not multiples of $p=503$, that is, the number of $k$ such that the combinations $\displaystyle {2012\choose k}$ are multiples of 2012 is $1501-3=1498$.