Base | Representation |
---|---|
bin | 10110100011101001011… |
… | …110001110110111111101 |
3 | 12111012002111111012210020 |
4 | 112203221132032313331 |
5 | 200344103420240013 |
6 | 3144035235002353 |
7 | 216664024111416 |
oct | 26435136166775 |
9 | 5435074435706 |
10 | 1550105243133 |
11 | 548439aa9137 |
12 | 2105071393b9 |
13 | b3235c14475 |
14 | 5504dd5480d |
15 | 2a4c60cc623 |
hex | 168e978edfd |
1550105243133 has 4 divisors (see below), whose sum is σ = 2066806990848. Its totient is φ = 1033403495420.
The previous prime is 1550105243131. The next prime is 1550105243153. The reversal of 1550105243133 is 3313425010551.
It is a semiprime because it is the product of two primes, and also a Blum integer, because the two primes are equal to 3 mod 4.
It is a cyclic number.
It is not a de Polignac number, because 1550105243133 - 21 = 1550105243131 is a prime.
It is a junction number, because it is equal to n+sod(n) for n = 1550105243094 and 1550105243103.
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (1550105243131) 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, 258350873853 + ... + 258350873858.
It is an arithmetic number, because the mean of its divisors is an integer number (516701747712).
Almost surely, 21550105243133 is an apocalyptic number.
It is an amenable number.
1550105243133 is a deficient number, since it is larger than the sum of its proper divisors (516701747715).
1550105243133 is an equidigital number, since it uses as much as digits as its factorization.
1550105243133 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 516701747714.
The product of its (nonzero) digits is 27000, while the sum is 33.
The spelling of 1550105243133 in words is "one trillion, five hundred fifty billion, one hundred five million, two hundred forty-three thousand, one hundred thirty-three".
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