Base | Representation |
---|---|
bin | 1111010010010… |
… | …0101100011001 |
3 | 11110122022002121 |
4 | 3310210230121 |
5 | 112403112213 |
6 | 10210101241 |
7 | 1405645426 |
oct | 364445431 |
9 | 143568077 |
10 | 64113433 |
11 | 3321040a |
12 | 1957a821 |
13 | 1038a357 |
14 | 872cd4d |
15 | 596688d |
hex | 3d24b19 |
64113433 has 4 divisors (see below), whose sum is σ = 64885968. Its totient is φ = 63340900.
The previous prime is 64113419. The next prime is 64113457. The reversal of 64113433 is 33431146.
It is a happy number.
64113433 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.
It is a semiprime because it is the product of two primes, and also a Blum integer, because the two primes are equal to 3 mod 4.
It is a cyclic number.
It is not a de Polignac number, because 64113433 - 29 = 64112921 is a prime.
It is a super-2 number, since 2×641134332 = 8221064582090978, which contains 22 as substring.
It is a Duffinian number.
It is a junction number, because it is equal to n+sod(n) for n = 64113398 and 64113407.
It is not an unprimeable number, because it can be changed into a prime (64113493) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (13) of ones.
It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 386143 + ... + 386308.
It is an arithmetic number, because the mean of its divisors is an integer number (16221492).
Almost surely, 264113433 is an apocalyptic number.
It is an amenable number.
64113433 is a deficient number, since it is larger than the sum of its proper divisors (772535).
64113433 is an equidigital number, since it uses as much as digits as its factorization.
64113433 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 772534.
The product of its digits is 2592, while the sum is 25.
The square root of 64113433 is about 8007.0864239123. The cubic root of 64113433 is about 400.2361792709.
Adding to 64113433 its reverse (33431146), we get a palindrome (97544579).
It can be divided in two parts, 64113 and 433, that added together give a palindrome (64546).
The spelling of 64113433 in words is "sixty-four million, one hundred thirteen thousand, four hundred thirty-three".
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