Base | Representation |
---|---|
bin | 10011000110111010110… |
… | …111101111110010101011 |
3 | 11122112100012200121210122 |
4 | 103012322313233302223 |
5 | 133003212122414011 |
6 | 2443121354233455 |
7 | 163603563356525 |
oct | 23067267576253 |
9 | 4575305617718 |
10 | 1313100201131 |
11 | 4669789161a4 |
12 | 1925a272588b |
13 | 96a95180344 |
14 | 477a9358a15 |
15 | 2425406b3db |
hex | 131badefcab |
1313100201131 has 2 divisors, whose sum is σ = 1313100201132. Its totient is φ = 1313100201130.
The previous prime is 1313100201053. The next prime is 1313100201169. The reversal of 1313100201131 is 1311020013131.
It is a strong prime.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-1313100201131 is a prime.
It is a super-2 number, since 2×13131002011312 (a number of 25 digits) contains 22 as substring.
It is a Sophie Germain prime.
It is not a weakly prime, because it can be changed into another prime (1313100201431) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 656550100565 + 656550100566.
It is an arithmetic number, because the mean of its divisors is an integer number (656550100566).
Almost surely, 21313100201131 is an apocalyptic number.
1313100201131 is a deficient number, since it is larger than the sum of its proper divisors (1).
1313100201131 is an equidigital number, since it uses as much as digits as its factorization.
1313100201131 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 54, while the sum is 17.
Adding to 1313100201131 its reverse (1311020013131), we get a palindrome (2624120214262).
The spelling of 1313100201131 in words is "one trillion, three hundred thirteen billion, one hundred million, two hundred one thousand, one hundred thirty-one".
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