# Problem of the Month (August 2018)

The problem of packing squares of sides 1, 2, 3, ... n inside the smallest square has been well-studied. The problem of using squares 1, 2, 3, ... n to cover the largest square was studied in the September 2000 Math Magic. We generalize the problem even further. Given n squares of sides 1, 2, 3, ... n, what is the largest square that they can cover k times? In the process, what is the largest area of squares that can be unused and still achieve the covering?

Joe DeVincentis and Maurizio Morandi sent solutions.

The known solutions are shown below. The size of the square and the unnecessary squares are given.

Largest Square Covered k Times With Squares 1–n
n \ k123
1
1
00
2
2 {1}

1
0
3
3 {2,1}

2 {1}

1
4
4 {3,2,1}

3 {2,1}

2 {1}
5
6 {1} (MM)

4 {3,2,1}

3 {2,1}
6
9

6 {1} (MM)

4 {3,2,1}
7
11 {2}

7 {5}

6 {1} (MM)
8
13 {4,1}

8 {7,2,1}

7 {5}
9
16 {2,1}

10 {1} (MM)

8 {7,2,1}
10
18 {7,1} (MM)

12 {2,1} (MM)

10 {1}
11
21 {3} (MM)

14 {4} (MM)

11 {6,1}
12
24 {3,2,1} (MM)

16 {6,2}

12 {11,2,1}
13
28 {1}

19 {2,1}

14 {1}
14
31 {4,1} (MM)

21 {6,1} (MM)

16 {2,1}
15
34 {5,4,1} (MM)

24 {2,1} (MM)

18 {5}
16
38 {5,2,1} (MM)

26 {8,1} (JD)

20 {6,2}
17
41 {9,1} (MM)

29 {1} (JD)

22 {7,4,1}
18
45 {8,1} (MM)

31 {8,1} (JD)

25 {1} (MM)
19
49 {5} (MM)

34 {5,4,1} (MM)

27 {7,1} (MM)
20
53 {4,3,1} (MM)

36 {13,1} (MM)

29 {8,6,1} (MM)
21
57 {5,1} (MM)

39 {8,7,3} (MM)

32 {3} (MM)
22
61 {7,1} (MM)

42 {9} (JD)

34 {10,1} (MM)
23
65 {6,4,2,1} (MM)

45 {3} (JD)

36 {12,7,1} (JD)
24
69 {8,4,3,2} (MM)

47 {14,2,1} (JD)

39 {6,5,4,1} (JD)

Maurizio Morandi also considered the problem of covering the largest possible squares and tans with tans of sides 1, 2, 3, ... n:

Largest Square Covered With Tans 1–n
n=1

s = 1/2
n=2

s = 3/2
n=3

s = (6+3√2)/4 = 2.560+
n=4

s = (11+3√2)/4 = 3.810+
n=5

s = (26+11√2)/8 = 5.194+
n=6

s = (62+27√2)/15 = 6.678+
n=7

s = (76+34√2)/15 = 8.272+
n=8

s = (98+315√2)/54 = 10.064+
n=9

s = (124+367√2)/54 = 11.907+
n=10

s = (338+329√2)/58 = 13.849+
n=11

s = (55+51√2)/8 = 15.890+
n=12

s = (110+75√2)/12 = 18.005+

Largest Tan Covered With Tans 1–n
n=1

s = 1
n=2

s = 3/√2 = 2.121+
n=3

s = (3+3√2)/2 = 3.621+
n=4

s = (6+11√2)/4 = 5.389+
n=5

s = (11+13√2)/4 = 7.346+
n=6

s = (27+28√2)/7 = 9.515+
n=7

s = (16+47√2)/7 = 11.781+
n=8

s = 10+3√2 = 14.242+
n=9

s = (92+197√2)/22 = 16.845+
n=10

s = (161+191√2)/22 = 19.596+

Largest Equilateral Triangle Covered With Equilateral Triangles 1–n
n=1

1
n=2

2 {1}
n=3

3 {2,1}
n=4

5
n=5

7 {1}
n=6

9 {2,1}
n=7

23/2 =11.5 (MM)
n=8

14 {1}
n=9

117/7 = 16.714+ (MM)
n=10

175/9 = 19.444+ (MM)
n=11

89/4 = 22.25 (MM)
n=12

201/8 = 25.125 (MM)

If you can extend any of these results, please e-mail me. Click here to go back to Math Magic. Last updated 8/1/18.