Base | Representation |
---|---|
bin | 11101110100010111100110… |
… | …011101101110010110100011 |
3 | 122012100001200122022002022211 |
4 | 131310113212131232112203 |
5 | 114142113011332003413 |
6 | 1142525503404542551 |
7 | 36423463365061531 |
oct | 3564274635562643 |
9 | 565301618262284 |
10 | 131142103000483 |
11 | 38871065304693 |
12 | 12860274b51a57 |
13 | 58238550114a3 |
14 | 2455451261151 |
15 | 102649148c83d |
hex | 7745e676e5a3 |
131142103000483 has 2 divisors, whose sum is σ = 131142103000484. Its totient is φ = 131142103000482.
The previous prime is 131142103000481. The next prime is 131142103000511. The reversal of 131142103000483 is 384000301241131.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 131142103000483 - 21 = 131142103000481 is a prime.
Together with 131142103000481, it forms a pair of twin primes.
It is not a weakly prime, because it can be changed into another prime (131142103000481) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 65571051500241 + 65571051500242.
It is an arithmetic number, because the mean of its divisors is an integer number (65571051500242).
Almost surely, 2131142103000483 is an apocalyptic number.
131142103000483 is a deficient number, since it is larger than the sum of its proper divisors (1).
131142103000483 is an equidigital number, since it uses as much as digits as its factorization.
131142103000483 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 6912, while the sum is 31.
The spelling of 131142103000483 in words is "one hundred thirty-one trillion, one hundred forty-two billion, one hundred three million, four hundred eighty-three", and thus it is an aban number.
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