Selected Problems from Open Contests

a) Both the step that involves increasing two of the numbers by $3$ and the step that involves decreasing one of the numbers by $6$ result in the sum of all three numbers being changed by $6$. Thus the remainder when the sum of the three numbers is divided by $6$ will always be the same regardless of the number of steps taken. But as the sums $0+1+2$ and $1+2+3$ give different remainders when divided by $6$, it is impossible to reach the required end situation from the given initial situation.

b) First increase the second and the third numbers three times by $4\frac{1}{2}$; we end up with $0$, $14\frac{1}{2}$, $15\frac{1}{2}$ in the vertices. Now increase the first and the second numbers by $4\frac{1}{2}$ and also increase the first and the third numbers by $4\frac{1}{2}$; so we end up with $9$, $19$ and $20$ written in the three vertices, respectively. Finally decrease the first number once by $6$ and the other two three times by $6$, achieving the situation in question.

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Alternative solution.

a) Consider one of the numbers. The remainder when this number is divided by $3$ is the same regardless of the number of steps taken. Therefore, if we want to achieve the situation where $1$, $2$, $3$ are located in the three vertices, the numbers $1$ and $2$ should stay in the same vertices where they were at the beginning and $3$ has to be in the vertex where $0$ was. Notice that two increasings are exactly cancelled out by one decreasing. Thus, the vertices where the numbers remain the same should have undergone an even number of increasings and the vertex where $0$ is replaced by $3$ should have been exposed to an odd number of increasings. Hence there should have been an odd number of increasings in total which is impossible since each increasing step influences the numbers in two vertices.

b) As in Solution 1.