32. A number and its last digit are not divisible by 3 without a remainder. Prove that by appending its last digit to the number several times, you can obtain a number that is divisible by 3. What is the smallest number of times you need to append this digit? 2*
32. Since this number is not divisible by 3, the remainders obtained when dividing by 3 its last digit and the number represented by the remaining digits (let's call it the "shortened" number) are either the same, or the "shortened" number is divisible by 3 without a remainder. If the "shortened" number is divisible by 3, and the last digit of the given number, according to the condition, when divided by 3 gives remainders of 1 or 2, then it is sufficient to append this digit twice to the given number to obtain a number divisible by 3, since the sum of the remainders in this case will be $1+1+1=3$ or $2+2+2=6$. If, however, the "shortened" number when divided by 3 gives a remainder of 1 or 2, then the last digit when divided by 3 also gives the same respective remainders. In this case, it is sufficient to append the last digit once to the given number to obtain a number divisible by 3, since in this case the sum of the remainders will be $1+1+1=3$ or $2+2+2=6$. Thus, it is sufficient to append the last digit one or two times to the given number to obtain a number divisible by 3.