Base | Representation |
---|---|
bin | 1011111011110100110011… |
… | …0010111101001001001011 |
3 | 1201110111020120010102101221 |
4 | 2332331030302331021023 |
5 | 3204444200324442324 |
6 | 43524204211022511 |
7 | 2523030432330562 |
oct | 276751462751113 |
9 | 51414216112357 |
10 | 13122413515339 |
11 | 41aa203368697 |
12 | 157b26589aa37 |
13 | 742590bb106b |
14 | 3351b1b08bd9 |
15 | 17b5264830e4 |
hex | bef4ccbd24b |
13122413515339 has 2 divisors, whose sum is σ = 13122413515340. Its totient is φ = 13122413515338.
The previous prime is 13122413515231. The next prime is 13122413515439. The reversal of 13122413515339 is 93351531422131.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 13122413515339 - 235 = 13088053776971 is a prime.
It is a junction number, because it is equal to n+sod(n) for n = 13122413515295 and 13122413515304.
It is not a weakly prime, because it can be changed into another prime (13122413515439) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6561206757669 + 6561206757670.
It is an arithmetic number, because the mean of its divisors is an integer number (6561206757670).
Almost surely, 213122413515339 is an apocalyptic number.
13122413515339 is a deficient number, since it is larger than the sum of its proper divisors (1).
13122413515339 is an equidigital number, since it uses as much as digits as its factorization.
13122413515339 is an evil number, because the sum of its binary digits is even.
The product of its digits is 291600, while the sum is 43.
The spelling of 13122413515339 in words is "thirteen trillion, one hundred twenty-two billion, four hundred thirteen million, five hundred fifteen thousand, three hundred thirty-nine".
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