Gribamko A.v.

A line contains 2020 natural numbers. Each of them, starting from the third, is divisible by both the previous number and the sum of the two previous numbers.

What is the smallest value that the last number in the line can take?
Example. The condition of the problem is obviously satisfied by the numbers 1, 1, 2!, 3!, ..., 2019!, since for any natural $\$ \mathrm{k} \$$ the number $\$(\mathrm{k}+2)!$ is divisible both by $\$(\mathrm{k}+1)!$ and by $\$(\mathrm{k}+1)!+\mathrm{k}!=\mathrm{k}!(\mathrm{k}+2) \$$.

Estimate. Let \$a,b,c be three consecutive numbers in the sequence, but not the first three numbers. We will prove that \$1frac\{c\} \{b \}geqslant \frac $\{\mathrm{b}\}\{\mathrm{a}\}+1 \$$. By the condition $\$ \mid f r a c\{b\}\{\mathrm{a}\}=\mathrm{x} \$$, $\$$ frac $\{\mathrm{c}\}\{\mathrm{b}\}=\mathrm{y} \$$, where $\$ \mathrm{x} \$$ and \$y are natural numbers. Then $\$ c=b y=a x y \$$, and since \$c, it follows that \$xy \$ is divisible by \$x+1 \$ and since \$x

Igeqslant x + 1\$, which is what we needed to prove.

Notice that \$a_1, a_2 $\$ 2$, $\$$ \$ divides it and is not equal to it). From the above proof $\$$ a_ $4 \$ 2 \$ 3$ a_3\$, \$ we get,

that $\$$ a_k $\$ 2(\$ k=-1)$ ! for all natural $\$ k \$$.

## Answer

$2019!$.