Base | Representation |
---|---|
bin | 101110101110011110110… |
… | …1111101100010011011011 |
3 | 211201221100001121101121111 |
4 | 1131130331233230103123 |
5 | 1320204240034114211 |
6 | 21354122005121151 |
7 | 1231655305332334 |
oct | 135347557542333 |
9 | 24657301541544 |
10 | 6422012019931 |
11 | 2056618400162 |
12 | 8787669b91b7 |
13 | 37779316b1b1 |
14 | 182b8158128b |
15 | b20b8086421 |
hex | 5d73dbec4db |
6422012019931 has 2 divisors, whose sum is σ = 6422012019932. Its totient is φ = 6422012019930.
The previous prime is 6422012019887. The next prime is 6422012019937. The reversal of 6422012019931 is 1399102102246.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 6422012019931 - 213 = 6422012011739 is a prime.
It is a super-3 number, since 3×64220120199313 (a number of 39 digits) contains 333 as substring. Note that it is a super-d number also for d = 2.
It is not a weakly prime, because it can be changed into another prime (6422012019937) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 3211006009965 + 3211006009966.
It is an arithmetic number, because the mean of its divisors is an integer number (3211006009966).
Almost surely, 26422012019931 is an apocalyptic number.
6422012019931 is a deficient number, since it is larger than the sum of its proper divisors (1).
6422012019931 is an equidigital number, since it uses as much as digits as its factorization.
6422012019931 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 46656, while the sum is 40.
The spelling of 6422012019931 in words is "six trillion, four hundred twenty-two billion, twelve million, nineteen thousand, nine hundred thirty-one".
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