Let N be the number of integers k with 0≤ k≤100 for which binom100k is odd and also divisible by 3. Find N.

Odd: by Lucas' theorem (base 2), $\binom{100}{k}$ is odd iff the binary digits of $k$ are dominated by those of $100=1100100_2$. This has three 1-bits, so $2^3=8$ values of $k$: $0,4,32,36,64,68,96,100$.
Not divisible by 3: by Lucas (base 3), $100=10201_3$, and $\binom{100}{k}\not\equiv0\pmod3$ iff each base-3 digit of $k$ is at most the matching digit of $100$. Among the eight values above, only $k=0$ and $k=100$ satisfy this (for example $k=4=11_3$ has digit $1$ in the $3^1$ place where $100$ has $0$).
So $2$ of the $8$ odd values are not multiples of $3$, and $N=8-2=6$.