3. 

A robot performs the following two operations:

$1^{\circ}$ with a 2-denar coin, it adds 4 to each indicated number; $2^{\circ}$ with a 5-denar coin, it multiplies each indicated number by 3.

Is it possible, starting from the number 1, to obtain the number 2001 with a finite number of operations using the robot? If it is possible, with what minimum sum of denars can we achieve this?
Solution: Since $2001=3\cdot667$, it is enough to obtain the number 667.

Then from $667=3\cdot222+1=3\cdot221+4$ it follows that we need to obtain the number 221.

Similarly, we conclude further: $221=3\cdot73+2=3\cdot71+4+4, 71=3\cdot23+2=3\cdot21+4+4, 21=3\cdot7, 7=3\cdot1+4$. From this, it follows that starting from the number 1, with the two operations of the robot, we can obtain the number 2001 in the following way: $3\cdot1=3$ (operation $2^{\circ}$): $4+3=7$ (operation $1^{\circ}$); $3\cdot7=21; 3\cdot21=63; 4+63=67; 4+67=71; 3\cdot71=213$; $4+213=217; 4+217=221; 3\cdot221=663; 4+663=667; 3\cdot667=2001$. Thus, a total of 12 operations are needed, six of each, so for this, $12\cdot2+5$ denars are needed, i.e., 84 denars.

4

In triangle $\mathrm{ABC}$, the angle at vertex $\mathrm{A}$ is $75^{\circ}$

on side BC, point D is chosen such that triangles ABD and ACD are isosceles.