Anton, Artem, and Vera decided to solve 100 math problems together. Each of them solved 60 problems. We will call a problem difficult if it was solved by only one person, and easy if it was solved by all three. How much does the number of difficult problems differ from the number of easy ones?

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Draw intersecting circles, where each circle represents the set of problems solved by one of the friends.

## Solution

Let $\mathrm{S}_{1}$ be the number of problems solved only by Anton, $\mathrm{S}_{2}$ be the number of problems solved only by Artem, $\mathrm{S}_{3}$ be the number of problems solved only by Vera, $\mathrm{S}_{12}$ be the number of problems solved only by Anton and Artem, and so on. Then Anton solved $\mathrm{S}_{1}+\mathrm{S}_{12}+\mathrm{S}_{13}+\mathrm{S}_{123}=60$ problems, Artem solved $\mathrm{S}_{2}+\mathrm{S}_{12}+\mathrm{S}_{23}+\mathrm{S}_{123}=60$ problems, and Vera solved $\mathrm{S}_{3}+\mathrm{S}_{13}+\mathrm{S}_{23}+\mathrm{S}_{123}=60$ problems. The total number of problems is: $\mathrm{S}_{1}+\mathrm{S}_{2}+\mathrm{S}_{3}+\mathrm{S}_{12}+\mathrm{S}_{13}+\mathrm{S}_{23}+\mathrm{S}_{123}=100$. Adding the first three equations and subtracting the last one, multiplied by 2, we get:

$-\mathrm{S}_{1}-\mathrm{S}_{2}-\mathrm{S}_{3}+\mathrm{S}_{123}=-20$

This means that the number of difficult problems is 20 more than the number of easy problems, because $S_{1}+S_{2}+S_{3}$ is the number of difficult problems, and $S_{123}$ is the number of easy problems.