Base | Representation |
---|---|
bin | 10110111010100001… |
… | …10101101010011001 |
3 | 1011202100121010001212 |
4 | 23131100311222121 |
5 | 200143320112213 |
6 | 5352420552505 |
7 | 613600144631 |
oct | 133520655231 |
9 | 34670533055 |
10 | 12302113433 |
11 | 52432499a5 |
12 | 2473b52735 |
13 | 121092257c |
14 | 849bc16c1 |
15 | 4c004a9a8 |
hex | 2dd435a99 |
12302113433 has 2 divisors, whose sum is σ = 12302113434. Its totient is φ = 12302113432.
The previous prime is 12302113429. The next prime is 12302113441. The reversal of 12302113433 is 33431120321.
It is a weak prime.
It can be written as a sum of positive squares in only one way, i.e., 12167854864 + 134258569 = 110308^2 + 11587^2 .
It is a cyclic number.
It is not a de Polignac number, because 12302113433 - 22 = 12302113429 is a prime.
It is a junction number, because it is equal to n+sod(n) for n = 12302113399 and 12302113408.
It is not a weakly prime, because it can be changed into another prime (12302113133) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6151056716 + 6151056717.
It is an arithmetic number, because the mean of its divisors is an integer number (6151056717).
Almost surely, 212302113433 is an apocalyptic number.
It is an amenable number.
12302113433 is a deficient number, since it is larger than the sum of its proper divisors (1).
12302113433 is an equidigital number, since it uses as much as digits as its factorization.
12302113433 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 1296, while the sum is 23.
Adding to 12302113433 its reverse (33431120321), we get a palindrome (45733233754).
The spelling of 12302113433 in words is "twelve billion, three hundred two million, one hundred thirteen thousand, four hundred thirty-three".
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