Today is 31 day of the year.

31 is prime.

31 = 7 + 11 + 13, the sum of three consecutive primes.

31 is a Mersenne prime.

31 is the only known Mersenne emirp.

31 is a happy number.

31 is a primorial prime.

31 is the member of one primitive Pythagorean triple (31, 480, 481)

## Tuesday, 31 January 2012

## Monday, 30 January 2012

### Monday, 30 January 2012

Today is 30 day of the year.

30 = 2 x 3 x 5, which is the product of the first three primes.

This product makes 30;

30 = 13 + 17, the sum of two consecutive primes.

30 = 1 + 4 + 9 + 16, the sum of the first four squares.

The Aliquot sum of a number, n, is the sum of all the divisors of n. Thus the aliquot sum of 30 = 1 + 2 + 3 + 5 + 6 + 10 + 15 = 42. This then gives rise to an Aliquot sequence where a(n + 1) = aliquot sum of n. Starting with 30 we get a sequence consisting of the 16 numbers 30, 42, 54, 66, 78, 90, 144, 259, 45, 33, 15, 9, 4, 3, 1, 0.

30 = 2 x 3 x 5, which is the product of the first three primes.

This product makes 30;

- the smallest sphenic number.
- Primorial, 5#
- pronic, oblong, rectangular or heteromecic i.e. 5 x 6

30 = 13 + 17, the sum of two consecutive primes.

30 = 1 + 4 + 9 + 16, the sum of the first four squares.

The Aliquot sum of a number, n, is the sum of all the divisors of n. Thus the aliquot sum of 30 = 1 + 2 + 3 + 5 + 6 + 10 + 15 = 42. This then gives rise to an Aliquot sequence where a(n + 1) = aliquot sum of n. Starting with 30 we get a sequence consisting of the 16 numbers 30, 42, 54, 66, 78, 90, 144, 259, 45, 33, 15, 9, 4, 3, 1, 0.

## Sunday, 29 January 2012

### Sunday, 29 January 2012

Today is 29 day of the year.

29 is prime.

29 is the maximum number of pieces of pizza one can create from 7 cuts, 29 is a lazy caterer number.

29 is a member of two primitive, Pythagorean triples (20, 21, 29) and (29, 420, 421).

29 is part of Cunningham chain of length 2, e.g. 29, 59 because 29 is a Sophie Germain prime.

29 is a Lucas number.

29 is a Pillai prime because 18! + 1 = 29 * 220,771,507,094,069 but 29 is not one greater than 18.

29 is a primorial prime.

Primorials are analogous to factorials. The nth factorial is the product of the first n natural numbers

e.g. 5! = 5 x 4 x 3 x 2 x 1 = 120.

The nth primorial is the product of the first n primes

e.g. p5# = p1 x p2 x p3 x p4 x p5 = 2 x 3 x 5 x 7 x 11 = 2,310

A primorial prime is a prime that differs from a primorial by 1.

2 x 3 x 5 = 30 = 29 + 1

The 29 digit, palindromic number 20,431,106,772,402,320,427,760,113,402 can be expressed as the product of three pandigital numbers 2,067,945,831, 2,758,436,091 and 3,581,704,962. This was discovered by Frank Rubin in early January 2012.

29 is prime.

29 is the maximum number of pieces of pizza one can create from 7 cuts, 29 is a lazy caterer number.

29 is a member of two primitive, Pythagorean triples (20, 21, 29) and (29, 420, 421).

29 is part of Cunningham chain of length 2, e.g. 29, 59 because 29 is a Sophie Germain prime.

29 is a Lucas number.

29 is a Pillai prime because 18! + 1 = 29 * 220,771,507,094,069 but 29 is not one greater than 18.

29 is a primorial prime.

Primorials are analogous to factorials. The nth factorial is the product of the first n natural numbers

e.g. 5! = 5 x 4 x 3 x 2 x 1 = 120.

The nth primorial is the product of the first n primes

e.g. p5# = p1 x p2 x p3 x p4 x p5 = 2 x 3 x 5 x 7 x 11 = 2,310

A primorial prime is a prime that differs from a primorial by 1.

2 x 3 x 5 = 30 = 29 + 1

The 29 digit, palindromic number 20,431,106,772,402,320,427,760,113,402 can be expressed as the product of three pandigital numbers 2,067,945,831, 2,758,436,091 and 3,581,704,962. This was discovered by Frank Rubin in early January 2012.

## Saturday, 28 January 2012

### Any Number Minus its Reverse is Divisible by 9

Assume that we have the number x = a10^n + b.10^(n-1) + c10^(n-2) + ... + m10^1 + n10^0

where a, b, c...m, n members of the set {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}.

The number made up from the reverse of these digits y = n10^n + m10^(n-1) + ... + c10^2 + b10^1 + a10^0.

Without loss of generality we can assume that x > y, therefore

x - y = (a10^n + b.10^(n-1) + c10^(n-2) + ... + m10^1 + n10^0) - (n10^n + m10^(n-1) + ... + c10^2 + b10^1 + a10^0)

x - y = a10^n + b.10^(n-1) + c10^(n-2) + ... + m10^1 + n10^0 - n10^n - m10^(n-1) - ... - c10^2 - b10^1 - a10^0

x - y = a10^n - a10^0 + b10^(n-1) - b10^1 + c10^(n-2) - c10^2 + ... + m10^1 - m10^(n-1) + n10^0 - n10^n

x - y = a(10^n - 10^0) + b(10^(n-1) - b10^1) + c(10^(n-2) - 10^2) + ... + m(10^1 - 10^(n-1)) + n(10^0 - 10^n)

However, we know from before that any power of 10 minus a power of 10 is divisible by 9, thus each set of brackets of the form (10^a - 10^b) can be replace by 9ki for some ki member of the integers.

Therefore,

x - y = a9k1 + b9k2 + c9k3 + ... + m9kn-1 + n9kn

x - y = 9.(ak1 + bk2 + ck3 + ... + mkn-1 + nkn)

QED

Note

Without much effort it can easily be shown that any number minus a number made from a permutation of its digits must also be divisible by 9.

where a, b, c...m, n members of the set {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}.

The number made up from the reverse of these digits y = n10^n + m10^(n-1) + ... + c10^2 + b10^1 + a10^0.

Without loss of generality we can assume that x > y, therefore

x - y = (a10^n + b.10^(n-1) + c10^(n-2) + ... + m10^1 + n10^0) - (n10^n + m10^(n-1) + ... + c10^2 + b10^1 + a10^0)

x - y = a10^n + b.10^(n-1) + c10^(n-2) + ... + m10^1 + n10^0 - n10^n - m10^(n-1) - ... - c10^2 - b10^1 - a10^0

x - y = a10^n - a10^0 + b10^(n-1) - b10^1 + c10^(n-2) - c10^2 + ... + m10^1 - m10^(n-1) + n10^0 - n10^n

x - y = a(10^n - 10^0) + b(10^(n-1) - b10^1) + c(10^(n-2) - 10^2) + ... + m(10^1 - 10^(n-1)) + n(10^0 - 10^n)

However, we know from before that any power of 10 minus a power of 10 is divisible by 9, thus each set of brackets of the form (10^a - 10^b) can be replace by 9ki for some ki member of the integers.

Therefore,

x - y = a9k1 + b9k2 + c9k3 + ... + m9kn-1 + n9kn

x - y = 9.(ak1 + bk2 + ck3 + ... + mkn-1 + nkn)

QED

Note

Without much effort it can easily be shown that any number minus a number made from a permutation of its digits must also be divisible by 9.

### Saturday, 28 January 2012

Today is 28 day of the year.

28 = 2 ^2 x 7

28 = 2 + 3 + 5 + 7 + 11, i.e the sum of the first five primes.

28 = 1 + 2 + 4 + 7 + 14, i.e. the aliquot sum equals 12 which makes 12 a perfect number.

28 is a Keith number because a Fibonacci like sequence starting with 2 and 8 includes the number 28:

2, 8, 10, 18,

28 is a triangular number.

28 is a member of two primitive, Pythagorean triples (28, 45, 53) and (28, 195, 197)

28 is a happy number.

28 is a member of the Padovan sequence as was 21 exactly a week ago.

28 is a member of the Y-toothpick sequence. There are some nice illustrations here and here.

28 = 2 ^2 x 7

28 = 2 + 3 + 5 + 7 + 11, i.e the sum of the first five primes.

28 = 1 + 2 + 4 + 7 + 14, i.e. the aliquot sum equals 12 which makes 12 a perfect number.

28 is a Keith number because a Fibonacci like sequence starting with 2 and 8 includes the number 28:

2, 8, 10, 18,

**28**, 46 ...28 is a triangular number.

28 is a member of two primitive, Pythagorean triples (28, 45, 53) and (28, 195, 197)

28 is a happy number.

28 is a member of the Padovan sequence as was 21 exactly a week ago.

28 is a member of the Y-toothpick sequence. There are some nice illustrations here and here.

## Friday, 27 January 2012

### Any Power of 10 Minus a Power of 10 is Divisible by 9

Consider 10^a - 10^b where a, b, member of Natural numbers and a > b

Let k = 10^a - 10^b

k = 10^b.(10^(a-b) - 1)

but we know from before the any power of 10 minus 1 is divisible by 9

Assume that (10^(a-b) - 1) = 9c for some c member of natural numbers

k = 10^b.9c

Thus k is a multiple of 9.

QED

Let k = 10^a - 10^b

k = 10^b.(10^(a-b) - 1)

but we know from before the any power of 10 minus 1 is divisible by 9

Assume that (10^(a-b) - 1) = 9c for some c member of natural numbers

k = 10^b.9c

Thus k is a multiple of 9.

QED

### Friday, 27 January 2012

Today is 27 day of the year.

27 = 3^3

27 is the smaller of the two double digit cubes.

27 is a Smith number.

27 is a member of one primitive, Pythagorean triple (27, 364, 365).

27 is the sixth member of Flavius Josephus's sieve.

27 is the 34 and 37 member of Alcuin's sequence.

27 is the number of ways of mapping 3 points to 3 points.

27 = 3^3

27 is the smaller of the two double digit cubes.

27 is a Smith number.

27 is a member of one primitive, Pythagorean triple (27, 364, 365).

27 is the sixth member of Flavius Josephus's sieve.

27 is the 34 and 37 member of Alcuin's sequence.

27 is the number of ways of mapping 3 points to 3 points.

## Thursday, 26 January 2012

### Thursday, 26 January 2012

Today is 26 day of the year.

26 = 2 x 13

26 = 3 + 5 + 7 + 11, i.e. the sum of four consecutive primes.

26 is the fifth cake number, i.e. one can cut a cake into 26 pieces with five cuts.

26 is the sixth Eulerian number.

26 is the only natural number to be sandwiched between a square and a cube which implies that there is only one solution to x^2 + 1 = y^3 - 1where x and y are natural numbers; proof anyone?

If one considers the permutation of a set of four elements then some of the permutations are self-inverse. For example the permutation of the ordered set (a,b,c,d} into {c,d,a,b} can be described as the swapping of the first and third elements with a swapping of the fourth and second elements. If one was to repeat these two swaps on the set {c,d,a,b} then the set {a,b,c,d} would be the result. Thus the permutation is self-inverse.

The number self-inverse sets, or involutions, of five elements is 26.

26 = 2 x 13

26 = 3 + 5 + 7 + 11, i.e. the sum of four consecutive primes.

26 is the fifth cake number, i.e. one can cut a cake into 26 pieces with five cuts.

26 is the sixth Eulerian number.

26 is the only natural number to be sandwiched between a square and a cube which implies that there is only one solution to x^2 + 1 = y^3 - 1where x and y are natural numbers; proof anyone?

If one considers the permutation of a set of four elements then some of the permutations are self-inverse. For example the permutation of the ordered set (a,b,c,d} into {c,d,a,b} can be described as the swapping of the first and third elements with a swapping of the fourth and second elements. If one was to repeat these two swaps on the set {c,d,a,b} then the set {a,b,c,d} would be the result. Thus the permutation is self-inverse.

The number self-inverse sets, or involutions, of five elements is 26.

### Any Power of 10 Minus One is Divisible by 9

(10^x - 1) / 9 = y, x member of natural numbers implies y is a natural number.

Proof.

When x = 1 then y = (10^1 - 1) / 9 = (10 - 1) / 9 = 9 / 9 = 1

Assume that (10^a -1) / 9 = b for some a, b members of the natural numbers.

Now consider

c = (10^(a+1) - 1) / 9

c = (10^a.10 - 10 + 9) / 9

c = (10.(10^a - 1) + 9 ) / 9

c =10.(10^a - 1)/9 + 9/9

c = 10.b + 1

Since b is a natural number then so is c.

So, by induction, we have shown that (10^x - 1) / 9 is a natural number or, equivalently, 10^x - 1 is always divisible by 9.

Proof.

When x = 1 then y = (10^1 - 1) / 9 = (10 - 1) / 9 = 9 / 9 = 1

Assume that (10^a -1) / 9 = b for some a, b members of the natural numbers.

Now consider

c = (10^(a+1) - 1) / 9

c = (10^a.10 - 10 + 9) / 9

c = (10.(10^a - 1) + 9 ) / 9

c =10.(10^a - 1)/9 + 9/9

c = 10.b + 1

Since b is a natural number then so is c.

So, by induction, we have shown that (10^x - 1) / 9 is a natural number or, equivalently, 10^x - 1 is always divisible by 9.

## Wednesday, 25 January 2012

### Wednesday, 25 January 2012

Today is 25 day of the year.

25 = 5^2

25 is the smallest square that can be written as the sum of two squares.

25 is a member of two primitive, Pythagorean triples (7, 24, 25) and (25, 312, 313).

25 ends in the digit 5 which makes it an automorphic number.

25 has an Aliquot sum of 6 and 6, being perfect, has an Aliquot sum of 6 which makes 25 an Aspiring number.

25 is a lucky number.

25 is a Cullen number.

25 = 5^2

25 is the smallest square that can be written as the sum of two squares.

25 is a member of two primitive, Pythagorean triples (7, 24, 25) and (25, 312, 313).

25 ends in the digit 5 which makes it an automorphic number.

25 has an Aliquot sum of 6 and 6, being perfect, has an Aliquot sum of 6 which makes 25 an Aspiring number.

25 is a lucky number.

25 is a Cullen number.

## Tuesday, 24 January 2012

### Tuesday, 24 January 2012

Today is 24 day of the year.

24 = 2^3 x 3

24 = 11 + 13, i.e. the sum of two consecutive primes.

24 = 4 x 3 x 2 x 1 = 4! which means there are 24 ways of arranging the four letters a, b, c, and d.

24 is in two primitive, Pythagorean triplets (7, 24, 25) and (24, 143, 145).

24 is the eighth tribonacci number.

24 is the number of moves required to solve the a game of Frogs and Toads with four of each.

One definition of Alcuin's sequence is;

the nth member of the sequence, denoted a(n) = the number of triangles with integer sides and perimeter n.

Clearly n = 0, n = 1 and n = 2 have no solution.

n = 3 has the single solution of an equilateral triangle with sides of 1 unit.

n = 4 you would initially think has the solution 1, 1, 2 but a little further thought reveals this to be a straight line and not a triangle.

n = 5 has the single solution (1, 2, 2)

The first perimeter that gives rise to more than one solution is n = 7 with solutions of (1, 3, 3) and (2, 2, 3).

24 is 31 and 34 member of Alcuin's sequence.

There is a college named after Alcuin at the University of York.

The 24 solutions for 31 and 34 are:

1: (1, 15, 15) (2, 16, 16)

2: (2, 14, 15) (3, 15, 16)

3: (3, 13, 15) (4, 14, 16)

4: (3, 14, 14) (4, 15, 15)

5: (4, 12, 15) (5, 13, 16)

6: (4, 13, 14) (5, 14, 15)

7: (5, 11, 15) (6, 12, 16)

8: (5, 12, 14) (6, 13, 15)

9: (5, 13, 13) (6, 14, 14)

10: (6, 10, 15) (7, 11, 16)

11: (6, 11, 14) (7, 12, 15)

12: (6, 12, 13) (7, 13, 14)

13: (7, 9, 15) (8, 10, 16)

14: (7, 10, 14) (8, 11, 15)

15: (7, 11, 13) (8, 12, 14)

16: (7, 12, 12) (8, 13, 13)

17: (8, 8, 15) (9, 9, 16)

18: (8, 9, 14) (9, 10, 15)

19: (8, 10, 13) (9, 11, 14)

20: (8, 11, 12) (9, 12, 13)

21: (9, 9, 13) (10, 10, 14)

22: (9, 10, 12) (10, 11, 13)

23: (9, 11, 11) (10, 12, 12)

24: (10, 10, 11) (11, 11, 12)

24 = 2^3 x 3

24 = 11 + 13, i.e. the sum of two consecutive primes.

24 = 4 x 3 x 2 x 1 = 4! which means there are 24 ways of arranging the four letters a, b, c, and d.

24 is in two primitive, Pythagorean triplets (7, 24, 25) and (24, 143, 145).

24 is the eighth tribonacci number.

24 is the number of moves required to solve the a game of Frogs and Toads with four of each.

One definition of Alcuin's sequence is;

the nth member of the sequence, denoted a(n) = the number of triangles with integer sides and perimeter n.

Clearly n = 0, n = 1 and n = 2 have no solution.

n = 3 has the single solution of an equilateral triangle with sides of 1 unit.

n = 4 you would initially think has the solution 1, 1, 2 but a little further thought reveals this to be a straight line and not a triangle.

n = 5 has the single solution (1, 2, 2)

The first perimeter that gives rise to more than one solution is n = 7 with solutions of (1, 3, 3) and (2, 2, 3).

24 is 31 and 34 member of Alcuin's sequence.

There is a college named after Alcuin at the University of York.

The 24 solutions for 31 and 34 are:

1: (1, 15, 15) (2, 16, 16)

2: (2, 14, 15) (3, 15, 16)

3: (3, 13, 15) (4, 14, 16)

4: (3, 14, 14) (4, 15, 15)

5: (4, 12, 15) (5, 13, 16)

6: (4, 13, 14) (5, 14, 15)

7: (5, 11, 15) (6, 12, 16)

8: (5, 12, 14) (6, 13, 15)

9: (5, 13, 13) (6, 14, 14)

10: (6, 10, 15) (7, 11, 16)

11: (6, 11, 14) (7, 12, 15)

12: (6, 12, 13) (7, 13, 14)

13: (7, 9, 15) (8, 10, 16)

14: (7, 10, 14) (8, 11, 15)

15: (7, 11, 13) (8, 12, 14)

16: (7, 12, 12) (8, 13, 13)

17: (8, 8, 15) (9, 9, 16)

18: (8, 9, 14) (9, 10, 15)

19: (8, 10, 13) (9, 11, 14)

20: (8, 11, 12) (9, 12, 13)

21: (9, 9, 13) (10, 10, 14)

22: (9, 10, 12) (10, 11, 13)

23: (9, 11, 11) (10, 12, 12)

24: (10, 10, 11) (11, 11, 12)

## Monday, 23 January 2012

### Monday, 23 January 2012

Today is 23 day of the year.

23 is prime.

23 = 5 + 7 + 11, i.e. the sum of three consecutive primes.

23 is a happy number.

23 is a factorial prime.

23 is a Pillai prime.

23 is a Woodall or Riesel prime.

23 is the fifth Sophie Germain prime.

23 is the fourth safe prime.

23 is a member of the Cunningham chain 2, 5, 11, 23, 47.

23 is a toothpick number.

23 is the lowest score that cannot be achieved with a single dart.

23 is a member of primitive, Pythagorean triple (23, 264, 265).

23 = 1 * 2 + 2 * 3 + 3 * 5, i.e. 1 times the first prime + 2 times the second prime + 3 times the third prime. Apart from the trivial 2 = 1 times the fist prime I cannot see any other prime that can be constructed this way for primes less than or equal to 17389.

23 is prime.

23 = 5 + 7 + 11, i.e. the sum of three consecutive primes.

23 is a happy number.

23 is a factorial prime.

23 is a Pillai prime.

23 is a Woodall or Riesel prime.

23 is the fifth Sophie Germain prime.

23 is the fourth safe prime.

23 is a member of the Cunningham chain 2, 5, 11, 23, 47.

23 is a toothpick number.

23 is the lowest score that cannot be achieved with a single dart.

23 is a member of primitive, Pythagorean triple (23, 264, 265).

23 = 1 * 2 + 2 * 3 + 3 * 5, i.e. 1 times the first prime + 2 times the second prime + 3 times the third prime. Apart from the trivial 2 = 1 times the fist prime I cannot see any other prime that can be constructed this way for primes less than or equal to 17389.

## Sunday, 22 January 2012

### Sunday, 22 January 2012

Today is 22 day of the year.

22 = 2 x 11

22 is the first hoax number.

22 is the second Smith number.

22 is the fourth pentagonal number.

22 is the sixth lazy caterer number.

Albert Wilansky coined the term Smith number when he noticed the defining property in the phone number of his brother-in-law Harold Smith: 493-7775. He noticed that 4937775 = 3*5*5*65837 and that 4+9+3+7+7+7+5 = 3+5+5+(6+5+8+3+7) that is, the sum of the digits of the number equals the sum of the digits of the factors.

Incidentally, did you work out what the sums added up to?

22 = 2 x 11

22 is the first hoax number.

22 is the second Smith number.

22 is the fourth pentagonal number.

22 is the sixth lazy caterer number.

Albert Wilansky coined the term Smith number when he noticed the defining property in the phone number of his brother-in-law Harold Smith: 493-7775. He noticed that 4937775 = 3*5*5*65837 and that 4+9+3+7+7+7+5 = 3+5+5+(6+5+8+3+7) that is, the sum of the digits of the number equals the sum of the digits of the factors.

Incidentally, did you work out what the sums added up to?

## Saturday, 21 January 2012

### Saturday, 21 January

Today is 21 day of the year.

21 = 3 x 7

21 is a lucky number.

21 is the third octagonal number.

21 is the third term in the Look and Say sequence.

21 is the sixth triangular number.

21 is the sixth Jacobsthal number.

21 is the eighth Fibonacci number.

21 is the thirteenth number in the Padovan sequence.

21 is a member of two primitive, Pythagorean triples (21, 20, 29) and (21, 220, 221).

21 = 3 x 7

21 is a lucky number.

21 is the third octagonal number.

21 is the third term in the Look and Say sequence.

21 is the sixth triangular number.

21 is the sixth Jacobsthal number.

21 is the eighth Fibonacci number.

21 is the thirteenth number in the Padovan sequence.

21 is a member of two primitive, Pythagorean triples (21, 20, 29) and (21, 220, 221).

## Friday, 20 January 2012

### Friday, 20 January 2012

Today is 20 day of the year.

20 = 2^2 x 5

20 is a tetrahedral number or, equivalently, the sum of the first four triangular numbers.

20 is the fifth oblong, promic, pronic or heteromecic number.

20 is the number of possible paths from one corner of a 3-by-3 grid to the opposite corner.

20 is a member of two primitive, Pythagorean triples (20, 21, 29) and (20, 99, 101).

20 = 2^2 x 5

20 is a tetrahedral number or, equivalently, the sum of the first four triangular numbers.

20 is the fifth oblong, promic, pronic or heteromecic number.

20 is the number of possible paths from one corner of a 3-by-3 grid to the opposite corner.

20 is a member of two primitive, Pythagorean triples (20, 21, 29) and (20, 99, 101).

## Thursday, 19 January 2012

### Thursday, 19 January 2012

Today is 19 day of the year.

19 is prime.

19 is a happy number.

19 is the fifth member of Flavius Josephus' sieve.

19 is a member of the primitive, Pythagorean triple (19, 80,81)

19 is the smallest prime whose reversal is composite.

19 is strictly non-palindromic as can be seen from its representation in different bases:

2: 10011

3: 201 - In base 3 all digits are used, i.e. 19 is pandigital in base 3.

4: 103

5: 34

6: 31

7: 25

8: 23

9: 21

10: 19

11: 18

12: 17

13: 16

14: 15

15: 14

16: 13

17: 12

19 is the largest prime, p, that is pandigital in some base b > 1 and b < p - 1.

19 is prime.

19 is a happy number.

19 is the fifth member of Flavius Josephus' sieve.

19 is a member of the primitive, Pythagorean triple (19, 80,81)

19 is the smallest prime whose reversal is composite.

19 is strictly non-palindromic as can be seen from its representation in different bases:

2: 10011

3: 201 - In base 3 all digits are used, i.e. 19 is pandigital in base 3.

4: 103

5: 34

6: 31

7: 25

8: 23

9: 21

10: 19

11: 18

12: 17

13: 16

14: 15

15: 14

16: 13

17: 12

19 is the largest prime, p, that is pandigital in some base b > 1 and b < p - 1.

## Wednesday, 18 January 2012

### Wednesday, 18 January 2012

Today is 18 day of the year.

18 = 2 x 3^2

18 = 7 + 11 i.e. the sum of two consecutive primes.

18 is an abundant number.

18 is the fourth triangular matchstick number. All triangular matchstick numbers are 3 times the equivalent triangular number.

The difference between an emirp pair is divisible by 18.

Proof

2 is not an emirp therefore all emirps are odd.

Assume that x and y are an emir pair and x > y.

Since they are odd we know that x = 2a + 1, y = 2b + 1 with a, b as whole numbers and a > b.

Calculate the difference between x and y:

x - y = (2a + 1) - (2b + 1) = 2a + 1 -2b - 1 = 2a -2b = 2(a - b)

therefore this difference is divisible by 2.

Assume that x, y are a 4 digit emirp pair.

Let x = 1000a + 100b + 10c + d with a, b, c, d, all single digits.

Thus y = 1000d + 100c + 10b + a.

The difference between them is:

x - y = (1000a + 100b + 10c + d) - (1000d + 100c + 10b + a)

= 1000a + 100b + 10c + d - 1000d - 100c - 10b - a

= 1000a - a + 100b - 10b + 10c - 100c + d - 1000d

= 999a +90b -90c - 999d

= 9(111a + 10b -10c - 111d)

Therefore the difference is divisible by 9.

Since 9 and 2 are coprime we have shown that the difference between any four digit emirp pair must be divisible by 18.

It is left as an exercise for the reader to convince themselves that this proof can be extended to an emirp pair of n digits and thus all emirp pairs.

18 = 2 x 3^2

18 = 7 + 11 i.e. the sum of two consecutive primes.

18 is an abundant number.

18 is the fourth triangular matchstick number. All triangular matchstick numbers are 3 times the equivalent triangular number.

The difference between an emirp pair is divisible by 18.

Proof

2 is not an emirp therefore all emirps are odd.

Assume that x and y are an emir pair and x > y.

Since they are odd we know that x = 2a + 1, y = 2b + 1 with a, b as whole numbers and a > b.

Calculate the difference between x and y:

x - y = (2a + 1) - (2b + 1) = 2a + 1 -2b - 1 = 2a -2b = 2(a - b)

therefore this difference is divisible by 2.

Assume that x, y are a 4 digit emirp pair.

Let x = 1000a + 100b + 10c + d with a, b, c, d, all single digits.

Thus y = 1000d + 100c + 10b + a.

The difference between them is:

x - y = (1000a + 100b + 10c + d) - (1000d + 100c + 10b + a)

= 1000a + 100b + 10c + d - 1000d - 100c - 10b - a

= 1000a - a + 100b - 10b + 10c - 100c + d - 1000d

= 999a +90b -90c - 999d

= 9(111a + 10b -10c - 111d)

Therefore the difference is divisible by 9.

Since 9 and 2 are coprime we have shown that the difference between any four digit emirp pair must be divisible by 18.

It is left as an exercise for the reader to convince themselves that this proof can be extended to an emirp pair of n digits and thus all emirp pairs.

## Tuesday, 17 January 2012

### Tuesday, 17 January

Today is 17 day of the year.

17 is prime.

17 = 2 + 3 + 5 + 7 i.e the sum of the first four primes.

17 = 2^3 + 3^2. 17 is the only prime of the form p^q + q^p where p and q are both prime.

17 is the most random number.

17 is a member of one primitive, Pythagorean triple (8, 15, 17).

17 is the third Fermat number defined by the form 2^2^n + 1.

17 is the seventh Tribonacci number.

17 is the only prime that can be expressed as the sum of four consecutive primes

Proof

17 is the sum of the first four consecutive primes.

The first four consecutive primes are the only set of four consecutive primes that include the number 2.

2 is the only even prime.

Therefore, all other sets of four consecutive primes have the characteristic that they are all odd numbers.

All odd numbers are of the form 2a + 1.

Assume we have four primes that are 2a + 1, 2b + 1, 2c + 1 and 2d + 1.

Their sum s = 2a + 1 + 2b + 1 + 2c + 1 + 2d + 1

s = 2(a + b + c + d) + 4 = 2(a + b + c + d + 2)

Thus the sum of any four consecutive primes, apart from the first four, is even.

If the sum is even then it cannot be prime.

QED

17 is prime.

17 = 2 + 3 + 5 + 7 i.e the sum of the first four primes.

17 = 2^3 + 3^2. 17 is the only prime of the form p^q + q^p where p and q are both prime.

17 is the most random number.

17 is a member of one primitive, Pythagorean triple (8, 15, 17).

17 is the third Fermat number defined by the form 2^2^n + 1.

17 is the seventh Tribonacci number.

17 is the only prime that can be expressed as the sum of four consecutive primes

Proof

17 is the sum of the first four consecutive primes.

The first four consecutive primes are the only set of four consecutive primes that include the number 2.

2 is the only even prime.

Therefore, all other sets of four consecutive primes have the characteristic that they are all odd numbers.

All odd numbers are of the form 2a + 1.

Assume we have four primes that are 2a + 1, 2b + 1, 2c + 1 and 2d + 1.

Their sum s = 2a + 1 + 2b + 1 + 2c + 1 + 2d + 1

s = 2(a + b + c + d) + 4 = 2(a + b + c + d + 2)

Thus the sum of any four consecutive primes, apart from the first four, is even.

If the sum is even then it cannot be prime.

QED

## Monday, 16 January 2012

### Monday, 16 January 2012

Today is 16 day of the year.

16 = 2^4

16 = 1 + 3 + 5 + 7 i.e. the sum of the first 4 odd numbers.

16 is the maximum number of pieces of pizza one can create with 5 cuts i.e. 16 is the fifth lazy caterer number.

16 is a member of the primitive, Pythagorean triple (16, 63, 65).

16 is the smallest perfect square number whose reverse is prime.

16 is the fifth member of the Euler zigzag numbers, beautifully illustrated here.

16 is the number of arrangements of three items.

The 16 arrangements of the three letters a, b and c are:

1) {a}

2) {b}

3) {c}

4) {ab}

5) {ac}

6) {ba}

7) {bc}

8) {ca}

9) {cb}

10) {abc}

11) {acb}

12) {bac}

13) {bca}

14) {cab}

15) {cba}

16) {}

16 = 2^4

16 = 1 + 3 + 5 + 7 i.e. the sum of the first 4 odd numbers.

16 is the maximum number of pieces of pizza one can create with 5 cuts i.e. 16 is the fifth lazy caterer number.

16 is a member of the primitive, Pythagorean triple (16, 63, 65).

16 is the smallest perfect square number whose reverse is prime.

16 is the fifth member of the Euler zigzag numbers, beautifully illustrated here.

16 is the number of arrangements of three items.

The 16 arrangements of the three letters a, b and c are:

1) {a}

2) {b}

3) {c}

4) {ab}

5) {ac}

6) {ba}

7) {bc}

8) {ca}

9) {cb}

10) {abc}

11) {acb}

12) {bac}

13) {bca}

14) {cab}

15) {cba}

16) {}

## Sunday, 15 January 2012

### Sunday, 15 January 2012

Today is 15 day of the year.

15 = 3 x 5

15 = 2^0 + 2^1 + 2^2 + 2^3 = 2^4 -1

15 is a triangular number.

15 = 3 + 5 + 7, i.e. the sum of 3 consecutive prime numbers.

15 is the fourth cake number.

15 is a member of the primitive, Pythagorean triple (8, 15, 17).

15 is the fifth Toothpick number.

15 is the smallest emirpimes.

15 is a lucky number.

15 is the fourth Bell number.

15 is the number of distinct rhyme schemes for a poem of 4 lines.

The 15 rhyme schemes are:

aaaa

aaab

aaba

aabb

aabc

abaa

abab

abac

abba

abbb

abbc

abca

abcb

abcc

abcd

15 = 3 x 5

15 = 2^0 + 2^1 + 2^2 + 2^3 = 2^4 -1

15 is a triangular number.

15 = 3 + 5 + 7, i.e. the sum of 3 consecutive prime numbers.

15 is the fourth cake number.

15 is a member of the primitive, Pythagorean triple (8, 15, 17).

15 is the fifth Toothpick number.

15 is the smallest emirpimes.

15 is a lucky number.

15 is the fourth Bell number.

15 is the number of distinct rhyme schemes for a poem of 4 lines.

The 15 rhyme schemes are:

aaaa

aaab

aaba

aabb

aabc

abaa

abab

abac

abba

abbb

abbc

abca

abcb

abcc

abcd

15 is the number of moves required to solve the frogs and toads game with 3 of each.

The moves are:

FFF_TTT

FFFT_TT - Slide T3

FF_TFTT - Jump F3

F_FTFTT - Slide F2

FTF_FTT - Jump T3

FTFTF_T - Jump T2

FTFTFT_ - Slide T1

FTFT_TF - Jump F3

FT_TFTF - Jump F2

_TFTFTF - Jump F1

T_FTFTF - Slide T3

TTF_FTF - Jump T2

TTFTF_F - Jump T1

TTFT_FF - Slide F2

TT_TFFF - Jump F1

TTT_FFF - Slide T3

## Saturday, 14 January 2012

### Saturday, 14 January 2012

Today is 14 day of the year.

14 = 2 x 7

14 is a Catalan number.

14 is the largest number whereby half the numbers less than it are prime.

14 = 2 x 7

14 is a Catalan number.

14 is the largest number whereby half the numbers less than it are prime.

## Friday, 13 January 2012

### Friday, 13 January 2012

Today is 13 day of the year.

13 is prime.

13 is a happy number.

13 is a Fibonacci number.

13 is a Tribonacci number with the first three numbers of 0, 0, 1.

13 is the smallest emirp.

13 is a member of the primitive, Pythagorean triples (5, 12, 13) and (13, 84, 85).

13 is a lucky number.

13 is the number of ways three competitors can finish a race if there can be ties.

The thirteen ways are:

1<2<3

1<3<2

2<1<3

2<3<1

3<1<2

3<2<1

1=2<3

1<2=3

1=2=3

1=3<2

2<1=3

2=3<1

3<1=2

13 is prime.

13 is a happy number.

13 is a Fibonacci number.

13 is a Tribonacci number with the first three numbers of 0, 0, 1.

13 is the smallest emirp.

13 is a member of the primitive, Pythagorean triples (5, 12, 13) and (13, 84, 85).

13 is a lucky number.

13 is the number of ways three competitors can finish a race if there can be ties.

The thirteen ways are:

1<2<3

1<3<2

2<1<3

2<3<1

3<1<2

3<2<1

1=2<3

1<2=3

1=2=3

1=3<2

2<1=3

2=3<1

3<1=2

## Thursday, 12 January 2012

### Thursday, 12 January 2012

Today is 12 day of the year.

12 = 2^2 * 3

12 = 3 + 4 + 5 which makes it the third term of the sequence A027480.

12 = 5 + 7, i.e. the sum of two consecutive primes.

12 is a member of two primitive, Pythagorean triples (5, 12, 13) and (12, 35, 37).

Starting at 0, write the numbers down in a clockwise spiral. 12 is the third term in one of the diagonals.

42 43 44 45 46 47 48

41 20 21 22 23 24 25

40 19 6 7 8 9 26

39 18 5

38 17 4 3

37 16 15 14 13

36 35 34 33 32 31

This diagonal gives the sequence 0, 2, 12, 30, 56, 90 132 ... which has the formula 2n(2n - 1).

12 = 2^2 * 3

12 = 3 + 4 + 5 which makes it the third term of the sequence A027480.

12 = 5 + 7, i.e. the sum of two consecutive primes.

12 is a member of two primitive, Pythagorean triples (5, 12, 13) and (12, 35, 37).

Starting at 0, write the numbers down in a clockwise spiral. 12 is the third term in one of the diagonals.

42 43 44 45 46 47 48

41 20 21 22 23 24 25

40 19 6 7 8 9 26

39 18 5

*0*1 10 2738 17 4 3

*2*11 2837 16 15 14 13

*12*2936 35 34 33 32 31

*30*This diagonal gives the sequence 0, 2, 12, 30, 56, 90 132 ... which has the formula 2n(2n - 1).

## Wednesday, 11 January 2012

### Wednesday, 11 January 2012

Today is 11 day of the year.

11 is prime.

11 is the first non-trivial, palindromic prime.

11 is strictly, non-palindromic.

11 is a Sophie Germain prime and a safe prime, it is therefore a member of the Cunningham chain 2, 5, 11, 23, 47

11 is the fourth lazy caterer number.

11 is a member of the number of leaf nodes in a binary tree.

11 is prime.

11 is the first non-trivial, palindromic prime.

11 is strictly, non-palindromic.

11 is a Sophie Germain prime and a safe prime, it is therefore a member of the Cunningham chain 2, 5, 11, 23, 47

11 is the fourth lazy caterer number.

11 is a member of the number of leaf nodes in a binary tree.

## Tuesday, 10 January 2012

### Tuesday, 10 January

Today is 10 day of the year.

10 = 2 x 5

10 is one of only 5 numbers that are both a triangular and a tetrahedral number.

10 = 2 + 3 + 5 i.e. it is the sum of the first three prime numbers.

10 is a happy number.

There are 10 ways to choose 3 objects from 5 objects where order is not important.

Let the five objects be the letters A, B, C, D and E.

The 10 combinations of these letters are

A, B, C

A, B, D

A, B, E

A, C, D

A. C. E

A, D, E

B, C, D

B, C, E

B, D, E

C, D, E

10 = 2 x 5

10 is one of only 5 numbers that are both a triangular and a tetrahedral number.

10 = 2 + 3 + 5 i.e. it is the sum of the first three prime numbers.

10 is a happy number.

There are 10 ways to choose 3 objects from 5 objects where order is not important.

Let the five objects be the letters A, B, C, D and E.

The 10 combinations of these letters are

A, B, C

A, B, D

A, B, E

A, C, D

A. C. E

A, D, E

B, C, D

B, C, E

B, D, E

C, D, E

## Monday, 9 January 2012

### Monday, 9 January

Today is 9 day of the year.

9 = 3^2, so it is a perfect square.

9 is the smallest odd, non-prime or composite number.

The sum of the first 9 primes is 100.

9 is a lucky number.

9 is a powerful number.

9 = 3^2, so it is a perfect square.

9 is the smallest odd, non-prime or composite number.

The sum of the first 9 primes is 100.

9 is a lucky number.

9 is a powerful number.

## Sunday, 8 January 2012

### Sunday, 8 January

Today is 8 day of the year.

8 = 2^3.

8 = 3 + 5, i.e. the sum of two consecutive primes.

8 is a member of just one primitive, Pthagorean triplet, namely (8, 15, 17).

8 is the third cake number.

8 is the sixth Fibonacci number.

It is conjectured that the difference between the squares of any two odd primes is a multiple of 8.

Update:

This is no longer a conjecture since it is easy to prove, as follows:

Consider an odd number, x = 2k + 1 for some k >= 0

Then x^2 = (2k+1)^2 = (2k + 1)(2k + 1) = 4k^2 + 4k + 1 = 4(k^2 +k) + 1

Consider k^2 + k

If k is odd then k^2 is odd and, therefore, k^2 + k is even.

If k is even then k^2 is even and, therefore k^2 + k is even.

Thus k^2 + k = 2l for some l >= 0.

Which means x^2 = 4.2l + 1 = 8l + 1.

Assume that we have another odd number, y,then we can safely assume that y^2 = 2m + 1 for some m >= 0.

This the difference between the two odd numbers squared, assuming x > y,

x^2 - y^2 = (8l + 1) - (8m + 1) = 8l + 1 - 8m - 1 = 8l - 8m = 8(l - m).

Thus we have shown that the difference between two odd numbers squared is a multiple of 8. Since the odd prime numbers are a subset of the odd numbers then this result holds for them as well.

For a conjecture that uses this result see Pe's conjecture.

8 = 2^3.

8 = 3 + 5, i.e. the sum of two consecutive primes.

8 is a member of just one primitive, Pthagorean triplet, namely (8, 15, 17).

8 is the third cake number.

8 is the sixth Fibonacci number.

It is conjectured that the difference between the squares of any two odd primes is a multiple of 8.

Update:

This is no longer a conjecture since it is easy to prove, as follows:

Consider an odd number, x = 2k + 1 for some k >= 0

Then x^2 = (2k+1)^2 = (2k + 1)(2k + 1) = 4k^2 + 4k + 1 = 4(k^2 +k) + 1

Consider k^2 + k

If k is odd then k^2 is odd and, therefore, k^2 + k is even.

If k is even then k^2 is even and, therefore k^2 + k is even.

Thus k^2 + k = 2l for some l >= 0.

Which means x^2 = 4.2l + 1 = 8l + 1.

Assume that we have another odd number, y,then we can safely assume that y^2 = 2m + 1 for some m >= 0.

This the difference between the two odd numbers squared, assuming x > y,

x^2 - y^2 = (8l + 1) - (8m + 1) = 8l + 1 - 8m - 1 = 8l - 8m = 8(l - m).

Thus we have shown that the difference between two odd numbers squared is a multiple of 8. Since the odd prime numbers are a subset of the odd numbers then this result holds for them as well.

For a conjecture that uses this result see Pe's conjecture.

## Saturday, 7 January 2012

### Saturday, January 7

Today is 7 day of the year.

7 is prime.

7 is the third lazy caterer number.

Since 2^7 -1 = 128 - 1 = 127 is prime, then 7 is a Mersenne prime.

7 is the only prime number anywhere in the universe that is one less than a perfect cube.

7 is a happy number.

7! x 2 = 10,080 which is the number of minutes in 7 days or one week.

7 is prime.

7 is the third lazy caterer number.

Since 2^7 -1 = 128 - 1 = 127 is prime, then 7 is a Mersenne prime.

7 is the only prime number anywhere in the universe that is one less than a perfect cube.

7 is a happy number.

7! x 2 = 10,080 which is the number of minutes in 7 days or one week.

## Thursday, 5 January 2012

### Friday, January 6

Today is 6 day of the year.

6 = 2 x 3.

6 is the smallest perfect number.

6 is a triangular number.

6 = 3!

6 is a semiprime.

6 is an unhappy number, i.e. the sum of the squares of the digits results in the following sequence, 36, 45, 41, 17, 50, 25, 29, 85, 89, 145, 42, 20, 4, 16, 37, 58, 89 the repeated 89 shows that the sequence will never reach 1.

6 = 2 x 3.

6 is the smallest perfect number.

6 is a triangular number.

6 = 3!

6 is a semiprime.

6 is an unhappy number, i.e. the sum of the squares of the digits results in the following sequence, 36, 45, 41, 17, 50, 25, 29, 85, 89, 145, 42, 20, 4, 16, 37, 58, 89 the repeated 89 shows that the sequence will never reach 1.

### Thursday, 5 January 2012

Today is 5 day of the year.

5 is prime.

5 is a safe prime.

5 = 2 + 3, i.e. the sum of the first two primes.

5 is Fibonacci number.

5 is an untouchable number and conjectured to be the only odd one.

There are 5 platonic solids.

5 is the fourth Pell number.

The series of fractions formed by taking a Pell number as the denominator and the sum of the Pell number and its predecessor, gets arbitrarily close to the square root of two.

5 is prime.

5 is a safe prime.

5 = 2 + 3, i.e. the sum of the first two primes.

5 is Fibonacci number.

5 is an untouchable number and conjectured to be the only odd one.

There are 5 platonic solids.

5 is the fourth Pell number.

The series of fractions formed by taking a Pell number as the denominator and the sum of the Pell number and its predecessor, gets arbitrarily close to the square root of two.

## Wednesday, 4 January 2012

### Wednesday, 4 January 2012

Today is 4 day of the year.

4 = 2 x 2

4 is a square number.

4 is a tetrahedral number.

4 is the second cake number.

4 is the first brilliant number.

4 = 2 x 2

4 is a square number.

4 is a tetrahedral number.

4 is the second cake number.

4 is the first brilliant number.

## Tuesday, 3 January 2012

### Tuesday, 3 January 2012

Today is 3 day of the year.

3 is a prime number

3 is a triangular number.

3 is the fifth number in the Fibonacci sequence.

3 is the member of just one primitive, Pythagorean triple (3, 4, 5).

3^3 + 4^3 + 5^3 = 6^3.

3 is a lucky number.

3 is a prime number

3 is a triangular number.

3 is the fifth number in the Fibonacci sequence.

3 is the member of just one primitive, Pythagorean triple (3, 4, 5).

3^3 + 4^3 + 5^3 = 6^3.

3 is a lucky number.

## Monday, 2 January 2012

### Monday 2 January

Today is 2 day of the year.

2 is the only even prime number.

For a polyhedron the number of faces plus the number of vertices minus the number of edges equals 2.

There is no prime between n! + 2 and n! + n (I am not aware of a proof for this to which I can link).

e.g.

10! + 2 = 3,628,802, 10! + 10 = 3,628,810

3,628,802 = 2 x 23 x 78,887

3,628,803 = 3 x 17 x 71,153

3,628,804 = 2^2 x 53 x 17,117

3,628,805 = 5 x 293 x 2,477

3,628,806 = 2 x 3 x 604,801

3,628,807 = 7 x 13 x 39,877

3,628,808 = 2^3 x 453,601

3,628,809 = 3^2 x 191 x 2,111

3,628,810 = 2 x 5 x 19 x 71 x 269

A corollary of this is that there are gaps between primes of any size. If one wanted to find a gap of at least m, then there is one between (m + 2)! + 2 and (m + 2) ! + (m + 2) since this is the previous proposition with n = m + 2 and

((m + 2) ! + (m + 2)) - ((m + 2)! + 2)

= (m + 2)! - (m + 2)! + (m + 2) - 2

= m as required.

2 is the only even prime number.

For a polyhedron the number of faces plus the number of vertices minus the number of edges equals 2.

There is no prime between n! + 2 and n! + n (I am not aware of a proof for this to which I can link).

e.g.

10! + 2 = 3,628,802, 10! + 10 = 3,628,810

3,628,802 = 2 x 23 x 78,887

3,628,803 = 3 x 17 x 71,153

3,628,804 = 2^2 x 53 x 17,117

3,628,805 = 5 x 293 x 2,477

3,628,806 = 2 x 3 x 604,801

3,628,807 = 7 x 13 x 39,877

3,628,808 = 2^3 x 453,601

3,628,809 = 3^2 x 191 x 2,111

3,628,810 = 2 x 5 x 19 x 71 x 269

A corollary of this is that there are gaps between primes of any size. If one wanted to find a gap of at least m, then there is one between (m + 2)! + 2 and (m + 2) ! + (m + 2) since this is the previous proposition with n = m + 2 and

((m + 2) ! + (m + 2)) - ((m + 2)! + 2)

= (m + 2)! - (m + 2)! + (m + 2) - 2

= m as required.

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