Base | Representation |
---|---|
bin | 100101111011110110101… |
… | …0001001101010000010111 |
3 | 200110102200000110012110201 |
4 | 1023313231101031100113 |
5 | 1140410303131302413 |
6 | 15031101535001331 |
7 | 1045453142062006 |
oct | 113675521152027 |
9 | 20412600405421 |
10 | 5213776040983 |
11 | 173017171a1a4 |
12 | 70256b069247 |
13 | 2ba870499883 |
14 | 1404c32c2c3d |
15 | 909502da9dd |
hex | 4bded44d417 |
5213776040983 has 2 divisors, whose sum is σ = 5213776040984. Its totient is φ = 5213776040982.
The previous prime is 5213776040959. The next prime is 5213776041043. The reversal of 5213776040983 is 3890406773125.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 5213776040983 - 221 = 5213773943831 is a prime.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (5213776040903) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (23) of ones.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 2606888020491 + 2606888020492.
It is an arithmetic number, because the mean of its divisors is an integer number (2606888020492).
It is a 1-persistent number, because it is pandigital, but 2⋅5213776040983 = 10427552081966 is not.
Almost surely, 25213776040983 is an apocalyptic number.
5213776040983 is a deficient number, since it is larger than the sum of its proper divisors (1).
5213776040983 is an equidigital number, since it uses as much as digits as its factorization.
5213776040983 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 7620480, while the sum is 55.
The spelling of 5213776040983 in words is "five trillion, two hundred thirteen billion, seven hundred seventy-six million, forty thousand, nine hundred eighty-three".
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