All Questions
5
questions
32
votes
3
answers
5k
views
When is $\binom{n}{k}$ divisible by $n$?
Is there any way of determining if $\binom{n}{k} \equiv 0\pmod{n}$. Note that I am aware of the case when $n =p$ a prime. Other than that there does not seem to be any sort of pattern (I checked up ...
- 2,106
2
votes
1
answer
365
views
Efficient computation of $\sum_{i=1}^{i=\left \lfloor {\sqrt{N}} \right \rfloor}\left \lfloor \frac{N}{i^{2}} \right \rfloor$
I have tried to find a closed form but did not succeed but is there an efficient way to calculate the following expression
$\sum_{i=1}^{i=\left \lfloor {\sqrt{N}} \right \rfloor}\left \lfloor \frac{N}...
- 37
3
votes
1
answer
169
views
$1\cdot3\cdot5 \cdots (p - 2) = (-1)^{m+k+1} \pmod p$, and $2\cdot4\cdot6\cdots(p - 1) = ( -1)^{m +k} \pmod p$.
Prove that if $p$ is a prime having the form $4k + 3$, and if $m$ is the number of quadratic residues less than $\frac p2$, then we have
$$1\cdot3\cdot5 \cdots (p - 2) = (-1)^{m+k+1} \pmod p, \text{ ...
- 12.7k
5
votes
1
answer
205
views
Conjecture: $\binom{n}{k } \mod m =0$ for all $k=1,2,3,\dots,n-1$ only when $m $ is a prime number and $n$ is a power of $m$
While playing with Pascal's triangle, I observed that $\binom{4}{k } \mod 2 =0$ for $k=1,2,3$,and $\binom{8}{k } \mod 2 =0$ for $k=1,2,3,4,5,6,7$ This made me curious about the values of $n>1$ and ...
- 6,581
2
votes
0
answers
848
views
$a^{(b^c)} \mod m$ where $c$ can be very very large
I am trying to solve the following problem.
I need to find the value of
$$
a^{(b^x)} \bmod m
$$
where $a,b$ are integers and
$$
x = \pmatrix{n\\0}^2 + \pmatrix{n\\1}^2 + ... + \pmatrix{n\\n}^2 ...
- 33