Base | Representation |
---|---|
bin | 110001011000010100… |
… | …101110000000101101 |
3 | 12001212011011220202011 |
4 | 301120110232000231 |
5 | 1332041441341101 |
6 | 40205133201221 |
7 | 3554630515642 |
oct | 613024560055 |
9 | 161764156664 |
10 | 53021433901 |
11 | 20539243067 |
12 | a338930211 |
13 | 4ccca1a237 |
14 | 27cdb15dc9 |
15 | 15a4c6e751 |
hex | c5852e02d |
53021433901 has 2 divisors, whose sum is σ = 53021433902. Its totient is φ = 53021433900.
The previous prime is 53021433893. The next prime is 53021433911. The reversal of 53021433901 is 10933412035.
It is a weak prime.
It can be written as a sum of positive squares in only one way, i.e., 33617222500 + 19404211401 = 183350^2 + 139299^2 .
It is a cyclic number.
It is not a de Polignac number, because 53021433901 - 23 = 53021433893 is a prime.
It is a super-2 number, since 2×530214339012 (a number of 22 digits) contains 22 as substring.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (53021433911) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 26510716950 + 26510716951.
It is an arithmetic number, because the mean of its divisors is an integer number (26510716951).
Almost surely, 253021433901 is an apocalyptic number.
It is an amenable number.
53021433901 is a deficient number, since it is larger than the sum of its proper divisors (1).
53021433901 is an equidigital number, since it uses as much as digits as its factorization.
53021433901 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 9720, while the sum is 31.
Adding to 53021433901 its reverse (10933412035), we get a palindrome (63954845936).
The spelling of 53021433901 in words is "fifty-three billion, twenty-one million, four hundred thirty-three thousand, nine hundred one".
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