Base | Representation |
---|---|
bin | 11000011111100100… |
… | …00111001000011101 |
3 | 1020221102110210111211 |
4 | 30033302013020131 |
5 | 203412310242131 |
6 | 10012455344421 |
7 | 643601401462 |
oct | 141762071035 |
9 | 36842423454 |
10 | 13149696541 |
11 | 563872a703 |
12 | 266b982111 |
13 | 13173c5737 |
14 | 8ca5b5669 |
15 | 51e6715b1 |
hex | 30fc8721d |
13149696541 has 2 divisors, whose sum is σ = 13149696542. Its totient is φ = 13149696540.
The previous prime is 13149696499. The next prime is 13149696559. The reversal of 13149696541 is 14569694131.
13149696541 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 9849768516 + 3299928025 = 99246^2 + 57445^2 .
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-13149696541 is a prime.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (13149696941) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (17) of ones.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6574848270 + 6574848271.
It is an arithmetic number, because the mean of its divisors is an integer number (6574848271).
Almost surely, 213149696541 is an apocalyptic number.
It is an amenable number.
13149696541 is a deficient number, since it is larger than the sum of its proper divisors (1).
13149696541 is an equidigital number, since it uses as much as digits as its factorization.
13149696541 is an odious number, because the sum of its binary digits is odd.
The product of its digits is 699840, while the sum is 49.
The spelling of 13149696541 in words is "thirteen billion, one hundred forty-nine million, six hundred ninety-six thousand, five hundred forty-one".
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