Base | Representation |
---|---|
bin | 111010010111101001011010… |
… | …0000111001000000111101111 |
3 | 2111022212210120101002012001212 |
4 | 1310233102310013020013233 |
5 | 1014243412122024042431 |
6 | 5015543243130025035 |
7 | 213100421532240131 |
oct | 16457226407100757 |
9 | 2438783511065055 |
10 | 513423412330991 |
11 | 139657712381738 |
12 | 49700a91b7817b |
13 | 1906380a2b7716 |
14 | 90adb7bbc7051 |
15 | 3e554c408a72b |
hex | 1d2f4b41c81ef |
513423412330991 has 2 divisors, whose sum is σ = 513423412330992. Its totient is φ = 513423412330990.
The previous prime is 513423412330981. The next prime is 513423412331021. The reversal of 513423412330991 is 199033214324315.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 513423412330991 - 210 = 513423412329967 is a prime.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (513423412330981) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 256711706165495 + 256711706165496.
It is an arithmetic number, because the mean of its divisors is an integer number (256711706165496).
Almost surely, 2513423412330991 is an apocalyptic number.
513423412330991 is a deficient number, since it is larger than the sum of its proper divisors (1).
513423412330991 is an equidigital number, since it uses as much as digits as its factorization.
513423412330991 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 2099520, while the sum is 50.
The spelling of 513423412330991 in words is "five hundred thirteen trillion, four hundred twenty-three billion, four hundred twelve million, three hundred thirty thousand, nine hundred ninety-one".
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