Base | Representation |
---|---|
bin | 11001110100011100110101… |
… | …000110111010000101000011 |
3 | 112220001210200020001020101112 |
4 | 121310130311012322011003 |
5 | 104340443212010422042 |
6 | 1041302405123553535 |
7 | 32630054632336106 |
oct | 3164346506720503 |
9 | 486053606036345 |
10 | 113555531342147 |
11 | 332006a3068664 |
12 | 1089b9a582b8ab |
13 | 4b4931895765b |
14 | 200818d24273d |
15 | d1dc8ed40982 |
hex | 6747351ba143 |
113555531342147 has 2 divisors, whose sum is σ = 113555531342148. Its totient is φ = 113555531342146.
The previous prime is 113555531342089. The next prime is 113555531342177. The reversal of 113555531342147 is 741243135555311.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 113555531342147 - 216 = 113555531276611 is a prime.
It is a junction number, because it is equal to n+sod(n) for n = 113555531342095 and 113555531342104.
It is not a weakly prime, because it can be changed into another prime (113555531342177) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (23) of ones.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 56777765671073 + 56777765671074.
It is an arithmetic number, because the mean of its divisors is an integer number (56777765671074).
Almost surely, 2113555531342147 is an apocalyptic number.
113555531342147 is a deficient number, since it is larger than the sum of its proper divisors (1).
113555531342147 is an equidigital number, since it uses as much as digits as its factorization.
113555531342147 is an odious number, because the sum of its binary digits is odd.
The product of its digits is 3780000, while the sum is 50.
Adding to 113555531342147 its reverse (741243135555311), we get a palindrome (854798666897458).
The spelling of 113555531342147 in words is "one hundred thirteen trillion, five hundred fifty-five billion, five hundred thirty-one million, three hundred forty-two thousand, one hundred forty-seven".
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