Find the maximum possible value of $\frac{A}{\gcd(A,B)} \text{ where } A={100\choose k} \text{ and } B={100\choose k+3}$ such that $30\leq k\leq70$: 
For an integer $30\leq k\leq70$, let $M$ be the maximum possible value of 
  $$\frac{A}{\gcd(A,B)} \text{ where } A={100\choose k} \text{ and } B={100\choose k+3}$$
  Find $M \mod 100$.

Okay, so directly, lets come to my attempt: 
$$\gcd(a,b)=\frac{a.b}{[a,b]}$$ where $[.]$ denotes the LCM. And then we can apply differentiation to find out the min of the denominator. But what would be the LCM of the two?
 A: Slightly more generally, if $A = {n \choose k}$ and $B = {n \choose k+3}$, $0 \le k \le n-3$, we have 
$$ \dfrac{A}{B} = \dfrac{(k+1)(k+2)(k+3)}{(n-k)(n-k-1)(n-k-2)}$$
If this is $s/t$ in lowest terms, then $\gcd(A,B) = A/s = B/t$.
You want to maximize (in the case $n=100$ with $30 \le k \le 70$) $A/\gcd(A,B) = s$, i.e. the numerator of $A/B$.  Let $f(k) = (k+1)(k+2)(k+3)$ and 
$g(k) = (100-k)(99-k)(98-k)$, so $$s = \dfrac{f(k)}{\gcd(f(k),g(k)}$$
$(k+1)(k+2)(k+3)$ increases pretty rapidly with $k$, so we should look at large $k$'s.  On the other hand, we want to avoid common factors of $f(k)$ and $g(k)$ as much as possible.  Some common factors are unavoidable: both $f(k)$ and $g(k)$ must be divisible by $6$, because at least one of $k+1,k+2,k+3$ is divisible by $2$ and at least one is divisible by $3$, and similarly for $100-k,99-k,98-k$.  Can we make $\gcd(f(k), g(k)) = 6$?
To avoid having a common factor of $4$ we want to avoid $k \equiv 2$ or $3 (\text{mod } 4)$.  To avoid having a common factor of $5$, we want to avoid $k \equiv 3$ or $4 (\text{mod } 5)$.  This would rule out $k = 62, 63, 64, 66, 67, 68, 69, 70$.  $k = 65$ would be a good candidate, but 
$f(65)$ and $g(65)$ turn out to be divisible by $11$ and $17$.  This leads
us to $61$, where indeed $f(61) =  2^7 \times 3^2 \times 7 \times 31$ and 
$g(61) = 2 \times 3 \times 13 \times 19 \times 37$ have gcd $6$.  Clearly
$$ \dfrac{f(k)}{\gcd(f(k),g(k))} \le \dfrac{f(61)}{6} \ \text{for } 0 < k \le 60 $$ 
while for $62 \le k \le 70$ we have $\gcd(f(k),g(k)) \ge 12$ and $f(k) \le f(70) < 2 f(61)$.  Thus the maximum is 
$$ \dfrac{f(61)}{6} = 41664$$
A: According to Mathematica, the maximum is $M=41664$, occurring for $k=61$.
