Base | Representation |
---|---|
bin | 11000001100010111010110… |
… | …10101001100100000001011 |
3 | 20222101000010221202200100211 |
4 | 30012023223111030200023 |
5 | 23433122443230132001 |
6 | 305052220343141551 |
7 | 14130450420142411 |
oct | 1406135325144013 |
9 | 228330127680324 |
10 | 53201413130251 |
11 | 15a51657832701 |
12 | 5b729597448b7 |
13 | 238bb4426a099 |
14 | d1cd681512b1 |
15 | 623d56913351 |
hex | 3062eb54c80b |
53201413130251 has 4 divisors (see below), whose sum is σ = 53205755760504. Its totient is φ = 53197070500000.
The previous prime is 53201413130233. The next prime is 53201413130291. The reversal of 53201413130251 is 15203131410235.
It is a semiprime because it is the product of two primes.
It is a cyclic number.
It is not a de Polignac number, because 53201413130251 - 213 = 53201413122059 is a prime.
It is a Duffinian number.
It is not an unprimeable number, because it can be changed into a prime (53201413130231) 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, 2171296750 + ... + 2171321251.
It is an arithmetic number, because the mean of its divisors is an integer number (13301438940126).
Almost surely, 253201413130251 is an apocalyptic number.
53201413130251 is a deficient number, since it is larger than the sum of its proper divisors (4342630253).
53201413130251 is a wasteful number, since it uses less digits than its factorization.
53201413130251 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 4342630252.
The product of its (nonzero) digits is 10800, while the sum is 31.
Adding to 53201413130251 its reverse (15203131410235), we get a palindrome (68404544540486).
The spelling of 53201413130251 in words is "fifty-three trillion, two hundred one billion, four hundred thirteen million, one hundred thirty thousand, two hundred fifty-one".
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