Counting Multiples of 2012 in Combination Numbers

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In summary, to find the number of integers in a given range, you can use the formula (upper limit - lower limit + 1) as long as the range is evenly spaced and includes both the upper and lower limits. This formula can also be used for ranges that include negative numbers, and there is no limit to the size of the range. However, it cannot be used for non-numeric ranges.
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Find the number of integers $k$ in the set ${0,\,1,\,\cdots,\,2012}$ such that the combination number $\displaystyle {2012\choose k}=\dfrac{2012!}{k!(2012-k)!}$ is a multiple of 2012.
 
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Factorizing $2012=4\times 503$, we see that $p=503$ is a prime. If $k$ is not a multiple of $p$, then

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

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 $1501-3=1498$.
 

Related to Counting Multiples of 2012 in Combination Numbers

1. How do you find the number of integers between two given numbers?

To find the number of integers between two given numbers, you can use the formula (larger number - smaller number) + 1. This takes into account both the starting and ending numbers in the range.

2. Can negative numbers be included when finding the number of integers?

Yes, negative numbers can be included when finding the number of integers between two given numbers. The formula (larger number - smaller number) + 1 still applies, but you may need to adjust the order of the numbers to ensure the larger number is subtracted from the smaller number.

3. Is zero considered an integer when finding the number of integers?

Yes, zero is considered an integer when finding the number of integers between two given numbers. Zero is a whole number and is included in the set of integers.

4. How do you find the number of integers in a given set of numbers?

To find the number of integers in a given set of numbers, you can count the total number of whole numbers in the set. This includes positive and negative integers, as well as zero.

5. Can fractions or decimals be considered integers when finding the number of integers?

No, fractions and decimals cannot be considered integers when finding the number of integers. Integers are whole numbers without any fractional or decimal components.

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