Base | Representation |
---|---|
bin | 1110010010110001000001… |
… | …11000011101011000011111 |
3 | 11010021210012221212102010012 |
4 | 13021120200320131120133 |
5 | 13104431414304442310 |
6 | 150503131441433435 |
7 | 6422552340530402 |
oct | 711304070353037 |
9 | 133253187772105 |
10 | 31431122343455 |
11 | a01895617864a |
12 | 363768b33b87b |
13 | 146cc32ab2b63 |
14 | 7a93bb0b7d39 |
15 | 3978e035dc05 |
hex | 1c9620e1d61f |
31431122343455 has 32 divisors (see below), whose sum is σ = 40442752051200. Its totient is φ = 23365515244032.
The previous prime is 31431122343373. The next prime is 31431122343539. The reversal of 31431122343455 is 55434322113413.
It is a cyclic number.
It is not a de Polignac number, because 31431122343455 - 218 = 31431122081311 is a prime.
It is a super-2 number, since 2×314311223434552 (a number of 28 digits) contains 22 as substring.
It is a congruent number.
It is an unprimeable number.
It is a polite number, since it can be written in 31 ways as a sum of consecutive naturals, for example, 358664774 + ... + 358752396.
It is an arithmetic number, because the mean of its divisors is an integer number (1263836001600).
Almost surely, 231431122343455 is an apocalyptic number.
31431122343455 is a deficient number, since it is larger than the sum of its proper divisors (9011629707745).
31431122343455 is a wasteful number, since it uses less digits than its factorization.
31431122343455 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 141143.
The product of its digits is 518400, while the sum is 41.
Adding to 31431122343455 its reverse (55434322113413), we get a palindrome (86865444456868).
The spelling of 31431122343455 in words is "thirty-one trillion, four hundred thirty-one billion, one hundred twenty-two million, three hundred forty-three thousand, four hundred fifty-five".
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