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Problem of the Week
Problem B and Solution
Playing with Bricks

Problem

Saskia has the following five sizes of LegoTM bricks.

A 2 times 2 brick with width 2 and length 2. A 2 times 3 brick with width 2 and length 3. A 2 times 4 brick with width 2 and length 4.

A 2 times 8 brick with width 2 and length 8. A 2 times 10 brick with width 2 and length 10.

For each question below, assume that Saskia will never run out of bricks.

  1. Saskia wants to make a row of bricks that measures \(2 \times 16\). Which of her brick sizes can she use if all bricks used must be the same size?

  2. Saskia wants to make a rectangular frame of bricks that measures \(14 \times 20\) on the outside, as shown. Which of her brick sizes can she use if all bricks used must be the same size?

    A frame with outer dimensions of 14 by 20. The frame border has a width of 2.

  3. Saskia wants to make a rectangular frame of bricks that measures \(320 \times 420\) on the outside. What is the largest brick size that she can use if all bricks used must be the same size?

Solution

In our solutions, we call the smaller number in the brick size the width, and call the larger number the length. So a \(2 \times 4\) brick has a width of \(2\) and a length of \(4\).

  1. Saskia can make a row that measures \(2 \times 16\) by using eight \(2 \times 2\) bricks, four \(2 \times 4\) bricks, or two \(2 \times 8\) bricks.

    In general, since the width of the row is the same as the width of each brick, we can focus only on the lengths. If we want to place some number of the same brick in a row and have the total length equal \(16\), then the possible bricks are ones whose length is a factor of \(16\). These are \(2 \times 2\), \(2 \times 4\), and \(2 \times 8\).

  2. We can think of the rectangular frame as two \(2 \times 20\) rows on the top and bottom, and two \(2 \times 10\) rows on the sides. If we placed these rows end to end, they would create a row with length \(20+20+10+10=60\). Since \(8\) is the only length that is not a factor of \(60\), we know for sure that it is not possible to make the frame using only \(2 \times 8\) bricks.

    For the other brick sizes, we need to show that it is possible to place them around the frame. For each brick size, we start at the top-left corner and place bricks horizontally moving to the right. Once we can no longer place bricks horizontally, we start placing the bricks vertically, moving down. We continue in this clockwise fashion, until all bricks are placed around the frame. This method works for the \(2 \times 2\), \(2 \times 3\), \(2 \times 4\), and \(2 \times 10\) bricks, as shown.

    2 times 2 bricks placed around the frame. 2 times 3 bricks placed around the frame.

    2 times 4 bricks placed around the frame. 2 by 10 bricks placed around the frame.

    Thus, the possible brick sizes are \(2 \times 2\), \(2 \times 3\), \(2 \times 4\), and \(2 \times 10\).

  3. We can think of the rectangular frame as two \(2 \times 320\) rows on the sides, and two \(2 \times 416\) rows on the top and bottom. If we placed these rows end to end, they would create a row with length \(320+320+416+416=1472\).

    Since \(10\) is not a factor of \(1472\), we know it is not possible to make the frame using only \(2 \times 10\) bricks. However, \(8\) is a factor of \(1472\), so we will attempt to place the \(2 \times 8\) bricks around the frame. It helps to notice that \(8\) is a factor of both \(320\) and \(416\). Thus, we can make two \(2 \times 320\) rows and two \(2 \times 416\) rows out of \(2 \times 8\) bricks. Putting these four rows together, with the \(2 \times 320\) rows on the sides, and the \(2 \times 416\) rows on the top and bottom gives the desired frame. Thus, the largest brick size that Saskia can use is \(2 \times 8.\)