Base | Representation |
---|---|
bin | 100110101001001001101… |
… | …1111101001010000110011 |
3 | 200210201202111022211200101 |
4 | 1031102103133221100303 |
5 | 1144004014323432313 |
6 | 15143510425533231 |
7 | 1055465605122511 |
oct | 115222337512063 |
9 | 20721674284611 |
10 | 5311054124083 |
11 | 17684506598a4 |
12 | 719399344217 |
13 | 2c6aa40baaa6 |
14 | 1450b0a474b1 |
15 | 932455b15dd |
hex | 4d4937e9433 |
5311054124083 has 2 divisors, whose sum is σ = 5311054124084. Its totient is φ = 5311054124082.
The previous prime is 5311054124039. The next prime is 5311054124153. The reversal of 5311054124083 is 3804214501135.
5311054124083 is digitally balanced in base 3, because in such base it contains all the possibile digits an equal number of times.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 5311054124083 - 225 = 5311020569651 is a prime.
It is not a weakly prime, because it can be changed into another prime (5311054124023) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 2655527062041 + 2655527062042.
It is an arithmetic number, because the mean of its divisors is an integer number (2655527062042).
Almost surely, 25311054124083 is an apocalyptic number.
5311054124083 is a deficient number, since it is larger than the sum of its proper divisors (1).
5311054124083 is an equidigital number, since it uses as much as digits as its factorization.
5311054124083 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 57600, while the sum is 37.
The spelling of 5311054124083 in words is "five trillion, three hundred eleven billion, fifty-four million, one hundred twenty-four thousand, eighty-three".
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