Base | Representation |
---|---|
bin | 11001001110001100011110… |
… | …00100011101111001111011 |
3 | 21021101020111122102201102112 |
4 | 30213012033010131321323 |
5 | 24232202331112144303 |
6 | 313543254431322535 |
7 | 14453043412644224 |
oct | 1447061704357173 |
9 | 237336448381375 |
10 | 55463313006203 |
11 | 1674394a2a2663 |
12 | 62791b338044b |
13 | 24c4225069a4b |
14 | d9a61da0a24b |
15 | 662adc29e0d8 |
hex | 32718f11de7b |
55463313006203 has 4 divisors (see below), whose sum is σ = 55473559328448. Its totient is φ = 55453066683960.
The previous prime is 55463313006103. The next prime is 55463313006247. The reversal of 55463313006203 is 30260031336455.
It is a semiprime because it is the product of two primes.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-55463313006203 is a prime.
It is a super-2 number, since 2×554633130062032 (a number of 28 digits) contains 22 as substring.
It is a Duffinian number.
It is not an unprimeable number, because it can be changed into a prime (55463313006103) by changing a digit.
It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 5123153003 + ... + 5123163828.
It is an arithmetic number, because the mean of its divisors is an integer number (13868389832112).
Almost surely, 255463313006203 is an apocalyptic number.
55463313006203 is a deficient number, since it is larger than the sum of its proper divisors (10246322245).
55463313006203 is a wasteful number, since it uses less digits than its factorization.
55463313006203 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 10246322244.
The product of its (nonzero) digits is 583200, while the sum is 41.
The spelling of 55463313006203 in words is "fifty-five trillion, four hundred sixty-three billion, three hundred thirteen million, six thousand, two hundred three".
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