Base | Representation |
---|---|
bin | 110001011001001010… |
… | …001001011110110101 |
3 | 12001220010112111120112 |
4 | 301121022021132311 |
5 | 1332104033040203 |
6 | 40210353312405 |
7 | 3555160542131 |
oct | 613112113665 |
9 | 161803474515 |
10 | 53035440053 |
11 | 2054613a0a2 |
12 | a341565705 |
13 | 50028b23bb |
14 | 27d19203c1 |
15 | 15a60d96d8 |
hex | c592897b5 |
53035440053 has 4 divisors (see below), whose sum is σ = 53141724000. Its totient is φ = 52929156108.
The previous prime is 53035440049. The next prime is 53035440083. The reversal of 53035440053 is 35004453035.
53035440053 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 53035440053 - 22 = 53035440049 is a prime.
It is a Duffinian number.
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (53035440043) by changing a digit.
It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 53141225 + ... + 53142222.
It is an arithmetic number, because the mean of its divisors is an integer number (13285431000).
Almost surely, 253035440053 is an apocalyptic number.
It is an amenable number.
53035440053 is a deficient number, since it is larger than the sum of its proper divisors (106283947).
53035440053 is a wasteful number, since it uses less digits than its factorization.
53035440053 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 106283946.
The product of its (nonzero) digits is 54000, while the sum is 32.
Adding to 53035440053 its reverse (35004453035), we get a palindrome (88039893088).
The spelling of 53035440053 in words is "fifty-three billion, thirty-five million, four hundred forty thousand, fifty-three".
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