Base | Representation |
---|---|
bin | 10101000100100011… |
… | …11010100110010111 |
3 | 1002012101112120121000 |
4 | 22202101322212113 |
5 | 141132002240210 |
6 | 5110310554343 |
7 | 550223001225 |
oct | 124221724627 |
9 | 32171476530 |
10 | 11312540055 |
11 | 48856a7863 |
12 | 22386689b3 |
13 | 10b38c61ab |
14 | 7945c9b15 |
15 | 463228ac0 |
hex | 2a247a997 |
11312540055 has 32 divisors (see below), whose sum is σ = 21169670400. Its totient is φ = 5715808416.
The previous prime is 11312540033. The next prime is 11312540101. The reversal of 11312540055 is 55004521311.
11312540055 is a `hidden beast` number, since 113 + 1 + 2 + 540 + 0 + 5 + 5 = 666.
11312540055 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.
It is not a de Polignac number, because 11312540055 - 26 = 11312539991 is a prime.
It is a Harshad number since it is a multiple of its sum of digits (27).
It is a congruent number.
It is an unprimeable number.
It is a pernicious number, because its binary representation contains a prime number (17) of ones.
It is a polite number, since it can be written in 31 ways as a sum of consecutive naturals, for example, 2202609 + ... + 2207738.
It is an arithmetic number, because the mean of its divisors is an integer number (661552200).
Almost surely, 211312540055 is an apocalyptic number.
11312540055 is a gapful number since it is divisible by the number (15) formed by its first and last digit.
11312540055 is a deficient number, since it is larger than the sum of its proper divisors (9857130345).
11312540055 is a wasteful number, since it uses less digits than its factorization.
11312540055 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 4410380 (or 4410374 counting only the distinct ones).
The product of its (nonzero) digits is 3000, while the sum is 27.
The spelling of 11312540055 in words is "eleven billion, three hundred twelve million, five hundred forty thousand, fifty-five".
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