Honouring a Great Teacher

Row $n$ of Pascal's triangle has the form $1\ \ n\ \ \dots\ \ n\ \ 1$ where $\dots$ consists of numbers greater than $n$ . This proves, first of all, that $n > 1$ occurs at least twice, and second, that $n$ cannot occur more than a finite number of times.

Now $1820 = 13 \cdot 7 \cdot 5 \cdot 2^2 = 2 \cdot 5 \cdot 14 \cdot 13$ . This invites us to try

$1820 \cdot 4! = 2 \cdot (2^3 \cdot 3) \cdot 5 \cdot 14 \cdot 13 = 16\cdot 15 \cdot 14 \cdot 13.$

This shows that $1820 = \left(\begin{array}{c} 16 \\ 4 \end{array}\right).$

We see that 1820 occurs at least four times: twice in row 1820, and twice in row 16.

Finally, symmetry shows that the number of occurrences is odd unless $\left(\begin{array}{c} 2n \\ n \end{array}\right) = 1820$ . But this is not the case, since $\left(\begin{array}{c} 14 \\ 7 \end{array}\right) = 3432 > 1820$ .

Thus C and D are true and we submit the answer $\boxed{2}$ .