Base | Representation |
---|---|
bin | 110100101111011010… |
… | …1110110101000011111 |
3 | 101211100022021200000221 |
4 | 1221132311312220133 |
5 | 3323424120403112 |
6 | 124010413502211 |
7 | 11116460611225 |
oct | 1513665665037 |
9 | 354308250027 |
10 | 113260325407 |
11 | 44040556919 |
12 | 19b4a7a1967 |
13 | a8acaca838 |
14 | 56a61d2b15 |
15 | 2e2d45ee07 |
hex | 1a5ed76a1f |
113260325407 has 2 divisors, whose sum is σ = 113260325408. Its totient is φ = 113260325406.
The previous prime is 113260325389. The next prime is 113260325429. The reversal of 113260325407 is 704523062311.
It is a weak prime.
It is an emirp because it is prime and its reverse (704523062311) is a distict prime.
It is a cyclic number.
It is not a de Polignac number, because 113260325407 - 211 = 113260323359 is a prime.
It is a super-2 number, since 2×1132603254072 (a number of 23 digits) contains 22 as substring.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (113260322407) 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, 56630162703 + 56630162704.
It is an arithmetic number, because the mean of its divisors is an integer number (56630162704).
Almost surely, 2113260325407 is an apocalyptic number.
113260325407 is a deficient number, since it is larger than the sum of its proper divisors (1).
113260325407 is an equidigital number, since it uses as much as digits as its factorization.
113260325407 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 30240, while the sum is 34.
Adding to 113260325407 its reverse (704523062311), we get a palindrome (817783387718).
The spelling of 113260325407 in words is "one hundred thirteen billion, two hundred sixty million, three hundred twenty-five thousand, four hundred seven".
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