Base | Representation |
---|---|
bin | 101101100000000000111011… |
… | …110101000001010101000011 |
3 | 222020112112100201012201000021 |
4 | 231200000323311001111003 |
5 | 202212114110210310201 |
6 | 1545334123330234311 |
7 | 60102423611552344 |
oct | 5540007365012503 |
9 | 866475321181007 |
10 | 200112120010051 |
11 | 58842089919387 |
12 | 1a53b0a0125997 |
13 | 8787651a617b2 |
14 | 375b69a5c20cb |
15 | 1820597352ea1 |
hex | b6003bd41543 |
200112120010051 has 2 divisors, whose sum is σ = 200112120010052. Its totient is φ = 200112120010050.
The previous prime is 200112120010043. The next prime is 200112120010073. The reversal of 200112120010051 is 150010021211002.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 200112120010051 - 23 = 200112120010043 is a prime.
It is not a weakly prime, because it can be changed into another prime (200112120020051) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 100056060005025 + 100056060005026.
It is an arithmetic number, because the mean of its divisors is an integer number (100056060005026).
Almost surely, 2200112120010051 is an apocalyptic number.
200112120010051 is a deficient number, since it is larger than the sum of its proper divisors (1).
200112120010051 is an equidigital number, since it uses as much as digits as its factorization.
200112120010051 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 40, while the sum is 16.
Adding to 200112120010051 its reverse (150010021211002), we get a palindrome (350122141221053).
The spelling of 200112120010051 in words is "two hundred trillion, one hundred twelve billion, one hundred twenty million, ten thousand, fifty-one".
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