Base | Representation |
---|---|
bin | 11011100010100011110101… |
… | …101101001001011000110011 |
3 | 120212212011120101001122122022 |
4 | 123202203311231021120303 |
5 | 111333431231200000201 |
6 | 1105334434233302055 |
7 | 34340535012615005 |
oct | 3342436555113063 |
9 | 525764511048568 |
10 | 121122200000051 |
11 | 35658700283888 |
12 | 1170236386492b |
13 | 5277a0b611243 |
14 | 21ca4baa9a375 |
15 | e009ee08e41b |
hex | 6e28f5b49633 |
121122200000051 has 2 divisors, whose sum is σ = 121122200000052. Its totient is φ = 121122200000050.
The previous prime is 121122199999981. The next prime is 121122200000089. The reversal of 121122200000051 is 150000002221121.
It is a strong prime.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-121122200000051 is a prime.
It is a junction number, because it is equal to n+sod(n) for n = 121122199999975 and 121122200000029.
It is not a weakly prime, because it can be changed into another prime (121122200006051) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 60561100000025 + 60561100000026.
It is an arithmetic number, because the mean of its divisors is an integer number (60561100000026).
Almost surely, 2121122200000051 is an apocalyptic number.
121122200000051 is a deficient number, since it is larger than the sum of its proper divisors (1).
121122200000051 is an equidigital number, since it uses as much as digits as its factorization.
121122200000051 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 80, while the sum is 17.
Adding to 121122200000051 its reverse (150000002221121), we get a palindrome (271122202221172).
The spelling of 121122200000051 in words is "one hundred twenty-one trillion, one hundred twenty-two billion, two hundred million, fifty-one", and thus it is an aban number.
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