Base | Representation |
---|---|
bin | 101101011001110101001… |
… | …011100100001101001011 |
3 | 102001021112222121102000212 |
4 | 231121311023210031023 |
5 | 402104442212114201 |
6 | 10345205413151335 |
7 | 441264201210305 |
oct | 55316513441513 |
9 | 12037488542025 |
10 | 3120112223051 |
11 | aa326101071a |
12 | 42484732b54b |
13 | 1982c104a435 |
14 | ab02b0d2975 |
15 | 562644b5abb |
hex | 2d6752e434b |
3120112223051 has 2 divisors, whose sum is σ = 3120112223052. Its totient is φ = 3120112223050.
The previous prime is 3120112223047. The next prime is 3120112223069. The reversal of 3120112223051 is 1503222110213.
It is a weak prime.
It is an emirp because it is prime and its reverse (1503222110213) is a distict prime.
It is a cyclic number.
It is not a de Polignac number, because 3120112223051 - 22 = 3120112223047 is a prime.
It is not a weakly prime, because it can be changed into another prime (3120112223021) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 1560056111525 + 1560056111526.
It is an arithmetic number, because the mean of its divisors is an integer number (1560056111526).
Almost surely, 23120112223051 is an apocalyptic number.
3120112223051 is a deficient number, since it is larger than the sum of its proper divisors (1).
3120112223051 is an equidigital number, since it uses as much as digits as its factorization.
3120112223051 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 720, while the sum is 23.
Adding to 3120112223051 its reverse (1503222110213), we get a palindrome (4623334333264).
The spelling of 3120112223051 in words is "three trillion, one hundred twenty billion, one hundred twelve million, two hundred twenty-three thousand, fifty-one".
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