Base | Representation |
---|---|
bin | 111110101111111111001… |
… | …111011110000010100101 |
3 | 120021020101001201022101111 |
4 | 332233333033132002211 |
5 | 1031122222404140143 |
6 | 13100544530225021 |
7 | 623353466000014 |
oct | 76577717360245 |
9 | 16236331638344 |
10 | 4312134443173 |
11 | 14128475266a7 |
12 | 597879125171 |
13 | 25382b8a1b1b |
14 | 10c9cbc98a7b |
15 | 7727dae499d |
hex | 3ebff3de0a5 |
4312134443173 has 2 divisors, whose sum is σ = 4312134443174. Its totient is φ = 4312134443172.
The previous prime is 4312134443171. The next prime is 4312134443209. The reversal of 4312134443173 is 3713444312134.
It is a weak prime.
It can be written as a sum of positive squares in only one way, i.e., 2304633682404 + 2007500760769 = 1518102^2 + 1416863^2 .
It is a cyclic number.
It is not a de Polignac number, because 4312134443173 - 21 = 4312134443171 is a prime.
Together with 4312134443171, it forms a pair of twin primes.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (4312134443171) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 2156067221586 + 2156067221587.
It is an arithmetic number, because the mean of its divisors is an integer number (2156067221587).
Almost surely, 24312134443173 is an apocalyptic number.
It is an amenable number.
4312134443173 is a deficient number, since it is larger than the sum of its proper divisors (1).
4312134443173 is an equidigital number, since it uses as much as digits as its factorization.
4312134443173 is an evil number, because the sum of its binary digits is even.
The product of its digits is 290304, while the sum is 40.
The spelling of 4312134443173 in words is "four trillion, three hundred twelve billion, one hundred thirty-four million, four hundred forty-three thousand, one hundred seventy-three".
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