Base | Representation |
---|---|
bin | 10011000110010001111… |
… | …111010000101110001001 |
3 | 11122110120102001111102012 |
4 | 103012101333100232021 |
5 | 133000311030231121 |
6 | 2442525333145305 |
7 | 163550564350031 |
oct | 23062177205611 |
9 | 4573512044365 |
10 | 1312414305161 |
11 | 46665672a423 |
12 | 192430a8b235 |
13 | 969b603b83b |
14 | 477422128c1 |
15 | 24213c32c5b |
hex | 13191fd0b89 |
1312414305161 has 2 divisors, whose sum is σ = 1312414305162. Its totient is φ = 1312414305160.
The previous prime is 1312414305119. The next prime is 1312414305221. The reversal of 1312414305161 is 1615034142131.
It is a weak prime.
It can be written as a sum of positive squares in only one way, i.e., 1123506721936 + 188907583225 = 1059956^2 + 434635^2 .
It is a cyclic number.
It is not a de Polignac number, because 1312414305161 - 210 = 1312414304137 is a prime.
It is not a weakly prime, because it can be changed into another prime (1312414305101) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 656207152580 + 656207152581.
It is an arithmetic number, because the mean of its divisors is an integer number (656207152581).
Almost surely, 21312414305161 is an apocalyptic number.
It is an amenable number.
1312414305161 is a deficient number, since it is larger than the sum of its proper divisors (1).
1312414305161 is an equidigital number, since it uses as much as digits as its factorization.
1312414305161 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 8640, while the sum is 32.
Adding to 1312414305161 its reverse (1615034142131), we get a palindrome (2927448447292).
The spelling of 1312414305161 in words is "one trillion, three hundred twelve billion, four hundred fourteen million, three hundred five thousand, one hundred sixty-one".
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