Base | Representation |
---|---|
bin | 11111010010101111001… |
… | …101010001011100100111 |
3 | 21121120121200112002222111 |
4 | 133102233031101130213 |
5 | 240213031310441020 |
6 | 4323520222541451 |
7 | 311235330046024 |
oct | 37225715213447 |
9 | 7546550462874 |
10 | 2150423140135 |
11 | 759a9788a426 |
12 | 2a8924437887 |
13 | 127a266978ba |
14 | 7611c50a54b |
15 | 3ae0dce1d5a |
hex | 1f4af351727 |
2150423140135 has 4 divisors (see below), whose sum is σ = 2580507768168. Its totient is φ = 1720338512104.
The previous prime is 2150423140109. The next prime is 2150423140153. The reversal of 2150423140135 is 5310413240512.
It is a semiprime because it is the product of two primes.
It is a cyclic number.
It is not a de Polignac number, because 2150423140135 - 213 = 2150423131943 is a prime.
It is a Duffinian number.
It is a junction number, because it is equal to n+sod(n) for n = 2150423140097 and 2150423140106.
It is a congruent number.
It is an unprimeable number.
It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 215042314009 + ... + 215042314018.
It is an arithmetic number, because the mean of its divisors is an integer number (645126942042).
Almost surely, 22150423140135 is an apocalyptic number.
2150423140135 is a deficient number, since it is larger than the sum of its proper divisors (430084628033).
2150423140135 is an equidigital number, since it uses as much as digits as its factorization.
2150423140135 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 430084628032.
The product of its (nonzero) digits is 14400, while the sum is 31.
Adding to 2150423140135 its reverse (5310413240512), we get a palindrome (7460836380647).
The spelling of 2150423140135 in words is "two trillion, one hundred fifty billion, four hundred twenty-three million, one hundred forty thousand, one hundred thirty-five".
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