Base | Representation |
---|---|
bin | 100110101000101010101… |
… | …0101100110110001011011 |
3 | 200210122001211002010220021 |
4 | 1031101111111212301123 |
5 | 1143444400343230201 |
6 | 15143215134411311 |
7 | 1055431013243005 |
oct | 115212525466133 |
9 | 20718054063807 |
10 | 5310011305051 |
11 | 1767a65a4a492 |
12 | 719148060b37 |
13 | 2c696903abcb |
14 | 145012351375 |
15 | 931d3c72ea1 |
hex | 4d455566c5b |
5310011305051 has 2 divisors, whose sum is σ = 5310011305052. Its totient is φ = 5310011305050.
The previous prime is 5310011305021. The next prime is 5310011305063. The reversal of 5310011305051 is 1505031100135.
It is a happy number.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 5310011305051 - 233 = 5301421370459 is a prime.
It is a super-2 number, since 2×53100113050512 (a number of 26 digits) contains 22 as substring.
It is not a weakly prime, because it can be changed into another prime (5310011305021) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 2655005652525 + 2655005652526.
It is an arithmetic number, because the mean of its divisors is an integer number (2655005652526).
Almost surely, 25310011305051 is an apocalyptic number.
5310011305051 is a deficient number, since it is larger than the sum of its proper divisors (1).
5310011305051 is an equidigital number, since it uses as much as digits as its factorization.
5310011305051 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 1125, while the sum is 25.
Adding to 5310011305051 its reverse (1505031100135), we get a palindrome (6815042405186).
The spelling of 5310011305051 in words is "five trillion, three hundred ten billion, eleven million, three hundred five thousand, fifty-one".
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