Base | Representation |
---|---|
bin | 1011010011000011101000… |
… | …1110100011010011010011 |
3 | 1121222112102002111220011211 |
4 | 2310300322032203103103 |
5 | 3112010300203010141 |
6 | 42230333120520551 |
7 | 2421314202334132 |
oct | 264607216432323 |
9 | 47875362456154 |
10 | 12422022313171 |
11 | 3a5a171833019 |
12 | 148757a180157 |
13 | 6c1512693163 |
14 | 30d32c88d919 |
15 | 1681d2da3d81 |
hex | b4c3a3a34d3 |
12422022313171 has 2 divisors, whose sum is σ = 12422022313172. Its totient is φ = 12422022313170.
The previous prime is 12422022313153. The next prime is 12422022313247. The reversal of 12422022313171 is 17131322022421.
12422022313171 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 12422022313171 - 239 = 11872266499283 is a prime.
It is not a weakly prime, because it can be changed into another prime (12422022310171) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6211011156585 + 6211011156586.
It is an arithmetic number, because the mean of its divisors is an integer number (6211011156586).
Almost surely, 212422022313171 is an apocalyptic number.
12422022313171 is a deficient number, since it is larger than the sum of its proper divisors (1).
12422022313171 is an equidigital number, since it uses as much as digits as its factorization.
12422022313171 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 8064, while the sum is 31.
Adding to 12422022313171 its reverse (17131322022421), we get a palindrome (29553344335592).
The spelling of 12422022313171 in words is "twelve trillion, four hundred twenty-two billion, twenty-two million, three hundred thirteen thousand, one hundred seventy-one".
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