6. Pete was given a new electric jigsaw on his birthday, with a feature to count the length of the cuts made. To try out the gift, Pete took a square piece of plywood with a side of 50 cm and cut it into squares with sides of 10 cm and 20 cm. How many squares in total were obtained, if the electric jigsaw shows a total cut length of $2 \mathbf{m} 80 \mathrm{~cm}$?
Answer: 16. Solution. The perimeter of the figures must be taken into account, in addition to the perimeter of the original square, from which we get that the total perimeter of the resulting squares is $280 \cdot 2 + 200 = 760$. Now, we can denote the number of squares through $x$ and $y$ respectively and solve the system $\left\{\begin{array}{c}10^{2} \cdot x + 20^{2} \cdot y = 50^{2}, \\ 4 \cdot 10 \cdot x + 4 \cdot 20 \cdot y = 760,\end{array}\right.$ or proceed with the following reasoning. If all the squares had a side of 10, there would be $6 \mathbf{~} 25$ of them, then the total perimeter would be $25 \cdot 40 = 1000$. If we replace four small squares $10 \times 10$ with one square $20 \times 20$, the perimeter would decrease by $160 - 80 = 80$ cm. To get 760 from 1000, this needs to be done 3 times. Therefore, there will be 3 squares $20 \times 20$ and 13 squares $10 \times 10$—a total of 16 squares.

## 2015/2016 Academic Year  CRITERIA FOR DETERMINING WINNERS AND PRIZE WINNERS ${ }^{2}$  of the Lomonosov Mathematics Olympiad  for school students  5-9 grades  ELIMINATION STAGE

WINNER:

From 91 points inclusive and above.

PRIZE WINNER:

From 70 points to 90 points inclusive.

## FINAL STAGE

WINNER (Diploma I degree):

Not to be awarded.

PRIZE WINNER (Diploma II degree):

100 points inclusive.

PRIZE WINNER (Diploma III degree):

From 80 points to 99 points inclusive.

[^0]:    ${ }^{2}$ Approved at the meeting of the jury of the Lomonosov Mathematics Olympiad for school students.