<s>[INST] Prove that the sum of numbers written on the board is always equal to $\binom{20}{2}$ We write the number $20$ as the sum of two numbers $a,b$ and write $ab$ on the board where $a,b \ge 1$. We do this again for $a,b$ until we get only (multiple instances of) the number $1$. Prove that the sum of numbers written on the board is always equal to $\binom{20}{2}$.
I don't know how should I work to get a combination; it is more likely to get a sum?
 [/INST] This process can be described as follows:
$F(n) = a \cdot (n-a) + F(a) + F(n-a)$ where $F(0) = F(1) = 0$.
(In other words, given some $n$, we're choosing a value for $a$ and letting $b=n-a$, so that $a+b=n$)
Try with $a=1$:
$F(n) = 1 \cdot (n - 1) + F(1) + F(n-1) = n - 1 + F(n-1) = \sum_{k=1}^{n} (k-1) = \frac{n(n-1)}{2}$
This at least gives us an easy way to compute $F(n)$ assuming we always choose $a=1$. 
Let's try again using induction, with the hypothesis that $F(n) = \frac{n(n-1)}{2}$ is true regardless of $a$ by substituting:
$F(n) = a \cdot (n-a) + \frac{a(a-1)}{2} + \frac{(n-a)(n-a-1)}{2} = \frac{n(n-1)}{2}$
All the $a$-terms cancel out, so $F(n)$ does not depend on $a$. It will always be equal to $\frac{n(n-1)}{2} = \binom{n}{2}$ 
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