55

55 = 5 × 11

55 = 1 × 2 + 3 + 4 × 5 + 6 + 7 + 8 + 9

55 = 9 + 8 + 7 + 6 + 5 × 4 + 3 + 2 × 1

55 = 0^6 + 1^8 − 2^9 − 3^7 + 4^5 + 5^0 + 6^4 + 7^3 + 8^1 + 9^2

55 = (111 − 1)/(1 + 1)
= 2 × 22 + 22/2
= 3³ + 3³ + 3/3
= 44 + 44/4
= 55
= 66 − 66/6
= 7 × 7 + 7 − 7/7
= 8 × 8 − 8 − 8/8
= (999 − 9)/(9 + 9)

55 = 1³ + 3³ + 3³

55 = 1² + 2² + 3² + 4² + 5²

55 = n * Prime(n) = 5*Prime(5)

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55 Take Sum of Cube of digits

5³+ 5³= 250

Repeat

2³ + 5³ + 0³= 133

Repeat

1³ + 3³ + 3³= 55

Number k such that 9^k + 2 is prime.

Integer k such that 10^k+21 is prime.

Fibonacci number: F(10)

55 = binomial(10 + 1, 2) is the 10th triangular number.

Centered icosahedral (or cuboctahedral) number, also crystal ball sequence for f.c.c. lattice.

55 cannot be written as a sum of 3 squares. (Integers that are not a sum of three squares)

The ring of integers of the field Q(sqrt(-55)) has class number 4.

Semiprime (Product of 2 Primes)

Factors: 1, 5, 11, 55

Fifty-five

Representations, Binary to Hexadecimal:

110111_2
2001_3
313_4
210_5
131_6
106_7
67_8
61_9
50_11
47_12
43_13
3d_14
3a_15
37_16

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