Radka received a large bag of lentils on the morning of her birthday. Every day after lunch, she would eat lentils, taking 3 lentils on weekdays and 5 lentils on Saturdays and Sundays. One evening, she realized she had eaten exactly 111 lentils. Which day of the week could Radka's birthday have been? Find both possibilities.

(E. Patáková)
Over a week, Radka ate $5 \cdot 3 + 2 \cdot 5 = 25$ lentils. Over four weeks, she ate 100 lentils, and 11 lentils remained. According to the conditions of the problem, this number needs to be expressed as the sum of several threes and fives, and this can be done in only one way: $11 = 2 \cdot 3 + 5$.

Radka, therefore, ate lentils for four whole weeks, two weekdays, and one weekend day. This means that Radka's indulgence ended either with a trio of days Thursday-Friday-Saturday, or Sunday-Monday-Tuesday. In the first case, Radka's birthday falls on a Thursday, and in the second case, on a Sunday.

Another solution. Similarly, we find that Radka ate 100 lentils over four weeks. We will examine all possible days when Radka could have had her birthday:

| birthday | amount of lentils eaten | conclusion |
| :---: | :---: | :---: |
| Monday | ..., 100 (no), 103 (Mon), 106 (Tue), 109 (Wed), 112 (Thu) | does not work |
| Tuesday | ..., 100 (Mon), 103 (Tue), 106 (Wed), 109 (Thu), 112 (Fri) | does not work |
| Wednesday | ..., 100 (Tue), 103 (Wed), 106 (Thu), 109 (Fri), 114 (Sat) | does not work |
| Thursday | ..., 100 (Wed), 103 (Thu), 106 (Fri), 111 (Sat) | works |
| Friday | ..., 100 (Thu), 103 (Fri), 108 (Sat), 113 (Sun) | does not work |
| Saturday | ..., 100 (Fri), 105 (Sat), 110 (Sun), 113 (Mon) | does not work |
| Sunday | ..., 100 (Sat), 105 (Sun), 108 (Mon), 111 (Tue) | works |

Radka's birthday could have been either on a Thursday or a Sunday.

Evaluation. 1 point for determining that Radka ate 25 lentils in a week; 1 point for determining that we can count 100 lentils this way; 2 points for the key idea of the solution (dividing $3+3+5$, proposing a table, etc.); 1 point for each correct solution.

If the solver finds and justifies one correct solution, award a maximum of 5 points. If the solver guesses one of the solutions, award 1 point.