Base | Representation |
---|---|
bin | 101110001010101001001001… |
… | …000110010100001010101011 |
3 | 222121220120121221221021002012 |
4 | 232022221021012110022223 |
5 | 203103113003121441324 |
6 | 1555455552051430135 |
7 | 60524156622550013 |
oct | 5612511106241253 |
9 | 877816557837065 |
10 | 203041510343339 |
11 | 5977146a3423a8 |
12 | 1a93297abb534b |
13 | 893a969a45505 |
14 | 381d393920c43 |
15 | 1871897ba920e |
hex | b8aa491942ab |
203041510343339 has 2 divisors, whose sum is σ = 203041510343340. Its totient is φ = 203041510343338.
The previous prime is 203041510343333. The next prime is 203041510343341. The reversal of 203041510343339 is 933343015140302.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 203041510343339 - 216 = 203041510277803 is a prime.
Together with 203041510343341, it forms a pair of twin primes.
It is a Chen prime.
It is a junction number, because it is equal to n+sod(n) for n = 203041510343296 and 203041510343305.
It is not a weakly prime, because it can be changed into another prime (203041510343333) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 101520755171669 + 101520755171670.
It is an arithmetic number, because the mean of its divisors is an integer number (101520755171670).
Almost surely, 2203041510343339 is an apocalyptic number.
203041510343339 is a deficient number, since it is larger than the sum of its proper divisors (1).
203041510343339 is an equidigital number, since it uses as much as digits as its factorization.
203041510343339 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 349920, while the sum is 41.
The spelling of 203041510343339 in words is "two hundred three trillion, forty-one billion, five hundred ten million, three hundred forty-three thousand, three hundred thirty-nine".
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