Base | Representation |
---|---|
bin | 1011111011001110000101… |
… | …0010111011010010000111 |
3 | 1201102111102012221021110021 |
4 | 2332303201102323102013 |
5 | 3204311404111204112 |
6 | 43515325010240011 |
7 | 2522212036155115 |
oct | 276634122732207 |
9 | 51374365837407 |
10 | 13112020022407 |
11 | 41a585a522609 |
12 | 1579244b86007 |
13 | 7415c5820047 |
14 | 3348a75d6ab5 |
15 | 17b118c99907 |
hex | bece14bb487 |
13112020022407 has 2 divisors, whose sum is σ = 13112020022408. Its totient is φ = 13112020022406.
The previous prime is 13112020022177. The next prime is 13112020022419. The reversal of 13112020022407 is 70422002021131.
It is a strong prime.
It is an emirp because it is prime and its reverse (70422002021131) is a distict prime.
It is a cyclic number.
It is not a de Polignac number, because 13112020022407 - 239 = 12562264208519 is a prime.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (13112020023407) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6556010011203 + 6556010011204.
It is an arithmetic number, because the mean of its divisors is an integer number (6556010011204).
Almost surely, 213112020022407 is an apocalyptic number.
13112020022407 is a deficient number, since it is larger than the sum of its proper divisors (1).
13112020022407 is an equidigital number, since it uses as much as digits as its factorization.
13112020022407 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 1344, while the sum is 25.
Adding to 13112020022407 its reverse (70422002021131), we get a palindrome (83534022043538).
The spelling of 13112020022407 in words is "thirteen trillion, one hundred twelve billion, twenty million, twenty-two thousand, four hundred seven".
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