Base | Representation |
---|---|
bin | 110100110100100101… |
… | …0001100111111000001 |
3 | 101211210101000101122021 |
4 | 1221221022030333001 |
5 | 3324302424311001 |
6 | 124035520041441 |
7 | 11123662156201 |
oct | 1515112147701 |
9 | 354711011567 |
10 | 113433432001 |
11 | 4411a24041a |
12 | 19b98763281 |
13 | a909929b86 |
14 | 56c11b4401 |
15 | 2e3d755ba1 |
hex | 1a6928cfc1 |
113433432001 has 2 divisors, whose sum is σ = 113433432002. Its totient is φ = 113433432000.
The previous prime is 113433431971. The next prime is 113433432011. The reversal of 113433432001 is 100234334311.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 80236227600 + 33197204401 = 283260^2 + 182201^2 .
It is an emirp because it is prime and its reverse (100234334311) is a distict prime.
It is a cyclic number.
It is not a de Polignac number, because 113433432001 - 219 = 113432907713 is a prime.
It is not a weakly prime, because it can be changed into another prime (113433432011) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 56716716000 + 56716716001.
It is an arithmetic number, because the mean of its divisors is an integer number (56716716001).
Almost surely, 2113433432001 is an apocalyptic number.
It is an amenable number.
113433432001 is a deficient number, since it is larger than the sum of its proper divisors (1).
113433432001 is an equidigital number, since it uses as much as digits as its factorization.
113433432001 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 2592, while the sum is 25.
Adding to 113433432001 its reverse (100234334311), we get a palindrome (213667766312).
The spelling of 113433432001 in words is "one hundred thirteen billion, four hundred thirty-three million, four hundred thirty-two thousand, one".
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