Base | Representation |
---|---|
bin | 110010000011101100… |
… | …111110100010000001 |
3 | 12010201212121221102012 |
4 | 302003230332202001 |
5 | 1340034244311223 |
6 | 40405252102305 |
7 | 3611645526545 |
oct | 620354764201 |
9 | 163655557365 |
10 | 53749213313 |
11 | 20882036554 |
12 | a500604995 |
13 | 50b7734943 |
14 | 285c641a25 |
15 | 15e8accc78 |
hex | c83b3e881 |
53749213313 has 2 divisors, whose sum is σ = 53749213314. Its totient is φ = 53749213312.
The previous prime is 53749213279. The next prime is 53749213387. The reversal of 53749213313 is 31331294735.
It is a weak prime.
It can be written as a sum of positive squares in only one way, i.e., 29821145344 + 23928067969 = 172688^2 + 154687^2 .
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-53749213313 is a prime.
It is a super-3 number, since 3×537492133133 (a number of 33 digits) contains 333 as substring.
It is a Sophie Germain prime.
It is a Curzon number.
It is not a weakly prime, because it can be changed into another prime (53749213393) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 26874606656 + 26874606657.
It is an arithmetic number, because the mean of its divisors is an integer number (26874606657).
Almost surely, 253749213313 is an apocalyptic number.
It is an amenable number.
53749213313 is a deficient number, since it is larger than the sum of its proper divisors (1).
53749213313 is an equidigital number, since it uses as much as digits as its factorization.
53749213313 is an evil number, because the sum of its binary digits is even.
The product of its digits is 204120, while the sum is 41.
The spelling of 53749213313 in words is "fifty-three billion, seven hundred forty-nine million, two hundred thirteen thousand, three hundred thirteen".
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