Problem 5. There is a set of two cards: 1 and 2. In one operation, it is allowed to form an expression using the numbers on the cards, arithmetic operations, and parentheses. If its value is a non-negative integer, it is issued on a new card. (For example, having cards 3, 5, and 7, you can form the expression 7 [ 5/3 and get a card with the number 25, or form the expression 3 哒 and get a card with the number 35.)

How to get a card with the number 2015 a) in 4 operations; b) in 3 operations $[8$ points] (I.V. Yashchenko)
Answer. a) For example

$[1+2=3 ; \quad 3+2=5 ; \quad 3-2-11=0 ; \quad 201[5=2015$

or

$[1+2=3 ; \quad[1]=[13 ; \quad 3[1=31 ; \quad([2+3) \cdot 13 \cdot 31=2015$.

b) $[1+2=3 ; \quad 3 \cdot 2-1=63 ; \quad(63+2) \cdot 31=2015$.

Comments. 1. To solve the problem, it is useful to first factorize 2015 into prime factors: $2015=5 \cdot 13 \cdot 31$.