Base | Representation |
---|---|
bin | 110100010010110011… |
… | …0111011111000011001 |
3 | 101201212101110211211111 |
4 | 1220211212323320121 |
5 | 3314442312112213 |
6 | 123331233100321 |
7 | 11053621133344 |
oct | 1504546737031 |
9 | 351771424744 |
10 | 112300113433 |
11 | 43698539402 |
12 | 199210b76a1 |
13 | a788bb1881 |
14 | 561488145b |
15 | 2dc3edca3d |
hex | 1a259bbe19 |
112300113433 has 2 divisors, whose sum is σ = 112300113434. Its totient is φ = 112300113432.
The previous prime is 112300113407. The next prime is 112300113463. The reversal of 112300113433 is 334311003211.
It is a weak prime.
It can be written as a sum of positive squares in only one way, i.e., 70393110489 + 41907002944 = 265317^2 + 204712^2 .
It is an emirp because it is prime and its reverse (334311003211) is a distict prime.
It is a cyclic number.
It is not a de Polignac number, because 112300113433 - 213 = 112300105241 is a prime.
It is a super-2 number, since 2×1123001134332 (a number of 23 digits) contains 22 as substring.
It is not a weakly prime, because it can be changed into another prime (112300113463) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 56150056716 + 56150056717.
It is an arithmetic number, because the mean of its divisors is an integer number (56150056717).
Almost surely, 2112300113433 is an apocalyptic number.
It is an amenable number.
112300113433 is a deficient number, since it is larger than the sum of its proper divisors (1).
112300113433 is an equidigital number, since it uses as much as digits as its factorization.
112300113433 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 648, while the sum is 22.
Adding to 112300113433 its reverse (334311003211), we get a palindrome (446611116644).
The spelling of 112300113433 in words is "one hundred twelve billion, three hundred million, one hundred thirteen thousand, four hundred thirty-three".
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