Some are left as puzzles for you to solve with a to show the solutions.

The facts below are all derived from Neil Sloane's On-line Encyclopedia of Integer Sequences where the sequences containing 2016 listed here are shown as, for instance,

A224950

which is also a link to the series itself.

Contents of this page

First some fun facts about the year 2016 and then about the number 2016.

A224950

There are 5 Mondays in February this year.
The last time that happened was 1988 and the next is 2044.

A135795

The transit of Mercury across the sun will be visible from somewhere on the earth this year.

The next time is in 2019.

A171466.

Small:
Large:
Very Large:

If we draw 64 points round a circle then join each to every other, there will be 64×63/2 = 2016 lines.
Here it is and it must surely be the prettiest way to see 2016!

10×9×8×7×6 | = 2016 |

5+4+3+2+1 |

2 | = 2, | 4×3 | = 4, | 6×5×4 | = 20, ... | 2n×(2n-1)×...×(n+2)×(n+1) |

1 | 2+1 | 3+2+1 | n+(n-1)+...+2+1 |

A110371

The next number with this pattern of a block of 1s followed by a block of 0s is 2032 which is 11111110000 in binary.

A006516, A043569

Also, the digit sum of the binary number 11111100000 is 6 which is a factor of 2016.

If we write 2016 *in every base from 2 to 10* and sum the digits each time, we will always get a total that is
itself a divisor of 2016.

Base | 2016 in base | Digitsum | 2016/Digitsum |
---|---|---|---|

2 | 11111100000 | 6 | 336 |

3 | 2202200 | 8 | 252 |

4 | 133200 | 9 | 224 |

5 | 31031 | 8 | 252 |

6 | 13200 | 6 | 336 |

7 | 5610 | 12 | 168 |

8 | 3740 | 14 | 144 |

9 | 2680 | 16 | 126 |

10 | 2016 | 9 | 224 |

for example, 2016 =

Base | 2016 in base | Digitsum | 2016/Digitsum |
---|---|---|---|

11 | 1 5 7 3 | 16 | 126 |

12 | 1 2 0 0 | 3 | 672 |

13 | 11 12 1 | 24 | 84 |

14 | 10 4 0 | 14 | 144 |

15 | 8 14 6 | 28 | 72 |

16 | 7 14 0 | 21 | 96 |

17 | 6 16 10 | 32 | 63 |

45^{2} − 3^{2} = 2016

46^{2} − 10^{2} = 2016

50^{2} − 22^{2} = 2016

54^{2} − 30^{2} = 2016

65^{2} − 47^{2} = 2016

71^{2} − 55^{2} = 2016

79^{2} − 65^{2} = 2016

90^{2} − 78^{2} = 2016

130^{2} − 122^{2} = 2016

171^{2} − 165^{2} = 2016

254^{2} − 250^{2} = 2016

505^{2} − 503^{2} = 2016

46

50

54

65

71

79

90

130

171

254

505

Now find three whole numbers a, b and c, all different, such that a

a=79, b=65, c=47

47^{2} = 2209, 65^{2} = 4225 and 79^{2}= 6241
have 2016 as the difference between one and the next.

47

A256418

How many ways can you write 32 as a sum of 7 squares if 0 and repetitions are allowed and also
each different order of the 7 squares counts as a different answer?
There are a total of 2016 solutions!

That is too many to list
by hand so try this simpler problem :

find the 6 ways of writing 32 as a sum of 7 squares, 0 and repetitions being allowed, if the
numbers to be squared are written in order of size. For example,

0^{2} + 0^{2} + 0^{2} + 0^{2} + 0^{2} + 4^{2} + 4^{2} = 32

but we do not count
0^{2} + 0^{2} + 0^{2} + 0^{2} + 4^{2} + 0^{2} + 4^{2} = 32

because 0,0,0,0,4,0,4 is not in the correct order.

Each of these is a basic solution:

0^{2} + 0^{2} + 0^{2} + 0^{2} + 0^{2} + 4^{2} + 4^{2} = 32

0^{2} + 0^{2} + 1^{2} + 1^{2} + 1^{2} + 2^{2} + 5^{2} = 32

0^{2} + 0^{2} + 1^{2} + 2^{2} + 3^{2} + 3^{2} + 3^{2} = 32

0^{2} + 0^{2} + 2^{2} + 2^{2} + 2^{2} + 2^{2} + 4^{2} = 32

0^{2} + 1^{2} + 1^{2} + 1^{2} + 2^{2} + 3^{2} + 4^{2} = 32

1^{2} + 1^{2} + 2^{2} + 2^{2} + 2^{2} + 3^{2} + 3^{2} = 32

A045849
0

0

0

0

0

1

Can you find 4 whole numbers a, b, c and d, all in arithmetic progression (when written in order each differs
from the next by the same amount) and the sum of their squares is 2016?

a=16, b=20, c=24, d=28

16^{2} + 20^{2} + 24^{2} + 28^{2} = 2016

16

A217843

2016 = 1 + 2 + 3 + ... + 63

2016 = 86 + 87 + ... + 106

2016 = 220 + 221 + ... + 228

2016 = 285 + 286 + ... + 291

2016 = 671 + 672 + 673

Runsums

2016 = 16 ×126: 2016 is the smallest multiple of 16 (after 16 itself) that ends 016

What is the next multiple of 16 ending **016**?

4016 = 251 × 16

2016 = 8 × 252

If we reverse the digits in 8 and 252, we get the same numbers. But also

2016 = 24 × 84 = 42 × 48

where if both factors are again reversed the new numbers still have the same product 2016!

So 2016 has two separate pairs of factors that, when the individual factors are reversed, they still have a product of 2016.

What is the smallest number with two pairs of factors like this?

What is the next such number after 2016?

The smallest number with two such pairs of factors is 484 and the next after 2016 is 2178:

484 = 11×44 = 22×22

2178 = 22×99 = 33×66

See also A262873. 484 = 11×44 = 22×22

2178 = 22×99 = 33×66

If we exclude the self-reversing factor pairs such as 8 × 252 which give the same pair when they are reversed, then see A228164

If we take a list of the numbers from 1 to 4, we can find 16 subsets of them:
{}, {1}, {2}, {3}, {4}, {1, 2}, {1, 3}, {1, 4}, {2, 3}, {2, 4}, {3, 4}, {1, 2, 3}, {1, 2, 4}, {1, 3, 4}, {2, 3, 4}, {1, 2, 3, 4}

Take each subset and multiply all the numbers in the subset. Some products will
be repeated but there are only 8 different products.

If we try this with the numbers from 1 to 14, there are exactly 2016 different products.

A060957

2^{15} = 32768 where the product of the digits is 3×2×7×6×8 = 2016.

A014257

What is the next year that is a product of the digits of a power of 2?

2700

2^{16} = 65536 and 6×5×5×3×6 = 2700

2

Once we have one such triangle, we can always multiply all the sides by 2 say to get 6-8-10 or by 3 to find 9-12-15 and so on. All these have the same

If we want

There is a whole series of PPTs where the longest side, the hypotenuse, is just 1 more than one of the other sides as here in the 3-4-5 triangle. The next is 5-12-13, and the series continues 7-24-25, 9-40-41, 11-60-61, 13-84-85, 15-112-113. If we add up the areas of these 7 triangles we have a total area of

The Pythagorean triangle 32-126-130 which has sides double those of the PPT 16-63-65 has an area 2016.

A083374

A055112: the areas

Pythagorean Triangles

Six of them are 52-80-84, 48-85-91, 58-70-96, 65-72-119, 32-126-130 (right-angled), 64-225-287.

In 64-225-287, the two sides 64=8

What are the other five integer-sided triangles with an area of 2016?

52-80-84 perimeter=216

48-85-91 perimeter=224

58-70-96 perimeter=224

65-72-119 perimeter=256

32-126-130 perimeter=288 Pythagorean

64-225-287 perimeter=576

60-112-164 perimeter=336

20-204-208 perimeter=432

35-192-221 perimeter=448

56-170-222 perimeter=448

109-148-255 perimeter=512

48-85-91 perimeter=224

58-70-96 perimeter=224

65-72-119 perimeter=256

32-126-130 perimeter=288 Pythagorean

64-225-287 perimeter=576

60-112-164 perimeter=336

20-204-208 perimeter=432

35-192-221 perimeter=448

56-170-222 perimeter=448

109-148-255 perimeter=512

A051586

A070150

A232461

side | 1 | 2 | 3 | 4 | 5 | 6 |
---|---|---|---|---|---|---|

shape | ||||||

size | 1 | 4 | 9 | 16 | 25 | 36 |

side | 1 | 2 | 3 | 4 | 5 | 6 |
---|---|---|---|---|---|---|

shape | ||||||

size | 1 | 3 | 6 | 10 | 15 | 21 |

- 2016 is 63
^{rd}triangular number: 1 + 2 + ... + 63 = 2016 - 2016 is the smallest Triangular number divisible by 32

A066561 - 2016 has 36 divisors and 36 is also a Triangular number. (The next number with this property is 3321 with 10 divisors.)

A116541

A076172 - 2016 is the 32
^{nd}hexagonal number: 1 + 6 + 15 + 28 + ... + 378 = 2016 - 2016 is the 14
^{th}24-gonal number - 2016 is the sixth 136-gonal number
- 2016 is the third 673-gonal number.

Hexagonal numbers: A000384,

24-gonal numbers: A051876.

See also Polygonal Numbers - Numbers as Shapes.

The first board here has 4 squares on its base. There are just 6 ways to place a domino on the squares: 4 with it horizontal and 2 verticallly.

If the pyramid board has a base of 32 squares and is 16 levels tall, we can place a domino on it in 2016 ways.

A000384

6 | 1 | 8 | 15 | |
---|---|---|---|---|

7 | 5 | 3 | 15 | |

2 | 9 | 4 | 15 | |

15↗ | 15 | 15 | 15 | ↖15 |

79 | 137 | 197 | 199 | 277 | 347 | 349 | 431 |

127 | 193 | 131 | 419 | 337 | 421 | 107 | 281 |

103 | 379 | 283 | 389 | 293 | 227 | 179 | 163 |

397 | 251 | 83 | 271 | 269 | 157 | 439 | 149 |

409 | 211 | 383 | 191 | 181 | 101 | 401 | 139 |

307 | 239 | 317 | 167 | 89 | 367 | 97 | 433 |

353 | 233 | 359 | 151 | 257 | 223 | 331 | 109 |

241 | 373 | 263 | 229 | 313 | 173 | 113 | 311 |

On the left is a magic square using the numbers 1 to 9 and each row, column and diagonal sums to 15.

We can use all 64 consecutive *prime numbers* between 79 and 439 to make an
8x8 magic square with magic constant 2016. This is the smallest magic constant in an 8x8 square of
consecutive primes.

A073520

A189188

Found by G Abe and A Suzuki, two "amateur" mathematicians and
reported in **The Study of Magic Squares** G Abe, 1957 (Japanese only).

See prime puzzles and
problems page.

There are 63 of size 1x1, some of size 2x2, and so on, up to the three of size 7x7.

For each size from 1x1 to 7x7, calculate how many squares there are of that size and therefore how long the perimeter is of all the squares of each size and so verify that the total is 2016.

There are | 9×7 | = | 63 | squares of size | 1×1 | each with perimeter | 4×1 | = | 4 | cm |

There are | 8×6 | = | 48 | squares of size | 2×2 | each with perimeter | 4×2 | = | 8 | cm |

There are | 7×5 | = | 35 | squares of size | 3×3 | each with perimeter | 4×3 | = | 12 | cm |

There are | 6×4 | = | 24 | squares of size | 4×4 | each with perimeter | 4×4 | = | 16 | cm |

There are | 5×3 | = | 15 | squares of size | 5×5 | each with perimeter | 4×5 | = | 20 | cm |

There are | 4×2 | = | 12 | squares of size | 6×6 | each with perimeter | 4×6 | = | 24 | cm |

There are | 3×1 | = | 3 | squares of size | 7×7 | each with perimeter | 4×7 | = | 28 | cm |

See also A083504

Click on the image

to enlarge or reduce

There are 2016 squares completely inside a circle of radius of 26.

A119677 are the alternate terms of A136485. Also see A261849.

A135273

Hypercubes are the continuation of the series:
a point, a line, a square, a cube into 4 and more dimensions.

Each is derived from the previous one by taking a copy of it but imagining that all the points leave a trail from the original to the copy, forming new edges in the new dimension.

The number of points will double each time we go to the next dimension.

The number of edges will double but also there are extra edges between the pairs points between the original and the copy.

The square faces and cubes are more difficult to see but the formula is as follows:

we have taken two copies of the previous dimensional object so we double the number of faces (cubes, ...) etc. But we also have made new edges, faces, etc equal to the number of 1-dimension-smaller objects, so the extra faces are made from each edge in the smaller dimension, the extra cubes are made form each face in the smaller dimension and so on.

So from a cube with 8 points and 6 square faces and 12 edges to the next dimension, we find the faces are 2×6 faces and 12 square faces from the 12 edges: a total of 24 square faces in the 4D hypercube.

Similarly, the number of edges in the 4D hypercube cube is twice 12 edges from the two cubes plus 8 more where the 8 points of the cube now become edge in the 4D version: a total of 24+8=32 edges.

Each is derived from the previous one by taking a copy of it but imagining that all the points leave a trail from the original to the copy, forming new edges in the new dimension.

Shape | ∙ | ||||
---|---|---|---|---|---|

Dimension | 0 | 1 | 2 | 3 | 4 |

The number of edges will double but also there are extra edges between the pairs points between the original and the copy.

The square faces and cubes are more difficult to see but the formula is as follows:

we have taken two copies of the previous dimensional object so we double the number of faces (cubes, ...) etc. But we also have made new edges, faces, etc equal to the number of 1-dimension-smaller objects, so the extra faces are made from each edge in the smaller dimension, the extra cubes are made form each face in the smaller dimension and so on.

So from a cube with 8 points and 6 square faces and 12 edges to the next dimension, we find the faces are 2×6 faces and 12 square faces from the 12 edges: a total of 24 square faces in the 4D hypercube.

Similarly, the number of edges in the 4D hypercube cube is twice 12 edges from the two cubes plus 8 more where the 8 points of the cube now become edge in the 4D version: a total of 24+8=32 edges.

Shape | Dimension | points | edges | faces | cubes | 4D | 5D | 6D | 7D | 8D | 9D |
---|---|---|---|---|---|---|---|---|---|---|---|

Point | 0 | 1 | |||||||||

Line | 1 | 2 | 1 | ||||||||

Square | 2 | 4 | 4 | 1 | |||||||

Cube | 3 | 8 | 12 | 6 | 1 | ||||||

4D Hypercube | 4 | 16 | 32 | 24 | 8 | 1 | |||||

5D Hypercube | 5 | 32 | 80 | 80 | 40 | 10 | 1 | ||||

6D Hypercube | 6 | 64 | 192 | 240 | 160 | 60 | 12 | 1 | |||

7D Hypercube | 7 | 128 | 448 | 672 | 560 | 280 | 84 | 14 | 1 | ||

8D Hypercube | 8 | 256 | 1024 | 1792 | 1792 | 1120 | 448 | 112 | 16 | 1 | |

9D Hypercube | 9 | 512 | 2304 | 4608 | 5376 | 4032 | 2016 | 672 | 144 | 18 | 1 |

© 1 January 2016 (updated 5 February 2016) Dr Ron Knott