1. The number 400 was divided into four parts such that if 1 is added to the first part, 2 is subtracted from the second part, the third part is multiplied by 3, and the fourth part is divided by 4, all the results will be equal. Into what parts was the number 400 divided?
Answer: $62, 65, 21$ and 252.

Solution. Let $4 x$ be the last (fourth) part of the number. After dividing it by 4, the result, equal to $x$, coincides with the third part multiplied by 3, so the third part of the number is $-\frac{x}{3}$. If the first part is increased by 1, and the second part is decreased by 2, their sum will be equal to $x+x=2 x$. Therefore, before the changes, this sum was $(x-1)+(x+2)=2 x+1$. Since the sum of all parts is 400, we have the equation

$$
2 x+1+\frac{x}{3}+4 x=400, \quad \text { from which } \quad x=63,
$$

that is, after changing all parts, the result turned out to be 63. Now it is easy to find the initial values of all parts: $62, 65, 21$ and 252.

Criteria. Only the answer - 2 points. Correctly formulated equation - 2 points. Solution with arithmetic errors - 4 points. Criteria are not cumulative. Full solution - 7 points.