Base | Representation |
---|---|
bin | 1011111010101100111001… |
… | …0010111011110101001001 |
3 | 1201101122102100122110020122 |
4 | 2332223032102323311021 |
5 | 3204140132444313303 |
6 | 43511253010033025 |
7 | 2521445212364321 |
oct | 276531622736511 |
9 | 51348370573218 |
10 | 13103111322953 |
11 | 41a1aa8820883 |
12 | 1577579578775 |
13 | 740805c58952 |
14 | 3342a038a281 |
15 | 17ac96b04e38 |
hex | beace4bbd49 |
13103111322953 has 2 divisors, whose sum is σ = 13103111322954. Its totient is φ = 13103111322952.
The previous prime is 13103111322937. The next prime is 13103111323087. The reversal of 13103111322953 is 35922311130131.
It is a weak prime.
It can be written as a sum of positive squares in only one way, i.e., 8240786438329 + 4862324884624 = 2870677^2 + 2205068^2 .
It is a cyclic number.
It is not a de Polignac number, because 13103111322953 - 24 = 13103111322937 is a prime.
It is not a weakly prime, because it can be changed into another prime (13103011322953) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6551555661476 + 6551555661477.
It is an arithmetic number, because the mean of its divisors is an integer number (6551555661477).
Almost surely, 213103111322953 is an apocalyptic number.
It is an amenable number.
13103111322953 is a deficient number, since it is larger than the sum of its proper divisors (1).
13103111322953 is an equidigital number, since it uses as much as digits as its factorization.
13103111322953 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 14580, while the sum is 35.
The spelling of 13103111322953 in words is "thirteen trillion, one hundred three billion, one hundred eleven million, three hundred twenty-two thousand, nine hundred fifty-three".
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