Base | Representation |
---|---|
bin | 11001110001000100010010… |
… | …111100010110010101000111 |
3 | 112212020120112210120222111022 |
4 | 121301010102330112111013 |
5 | 104323141031124234301 |
6 | 1041003514203301355 |
7 | 32604214215661433 |
oct | 3161042274262507 |
9 | 485216483528438 |
10 | 113323029914951 |
11 | 3312103307512a |
12 | 1086291951a85b |
13 | 4b30414c6446c |
14 | 1ddac148139c3 |
15 | d17bd3383d1b |
hex | 671112f16547 |
113323029914951 has 2 divisors, whose sum is σ = 113323029914952. Its totient is φ = 113323029914950.
The previous prime is 113323029914947. The next prime is 113323029914953. The reversal of 113323029914951 is 159419920323311.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 113323029914951 - 22 = 113323029914947 is a prime.
Together with 113323029914953, it forms a pair of twin primes.
It is a Chen prime.
It is a junction number, because it is equal to n+sod(n) for n = 113323029914893 and 113323029914902.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (113323029914953) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 56661514957475 + 56661514957476.
It is an arithmetic number, because the mean of its divisors is an integer number (56661514957476).
Almost surely, 2113323029914951 is an apocalyptic number.
113323029914951 is a deficient number, since it is larger than the sum of its proper divisors (1).
113323029914951 is an equidigital number, since it uses as much as digits as its factorization.
113323029914951 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 1574640, while the sum is 53.
The spelling of 113323029914951 in words is "one hundred thirteen trillion, three hundred twenty-three billion, twenty-nine million, nine hundred fourteen thousand, nine hundred fifty-one".
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