Base | Representation |
---|---|
bin | 1110111000100110001… |
… | …01100011011011000101 |
3 | 1210220001211110110011000 |
4 | 13130103011203123011 |
5 | 31334342330044031 |
6 | 1030535502003513 |
7 | 51643332406020 |
oct | 7342305433305 |
9 | 1726054413130 |
10 | 511421331141 |
11 | 187990010646 |
12 | 83149b76599 |
13 | 392c4444c96 |
14 | 1aa7804c1b7 |
15 | d48369b3e6 |
hex | 77131636c5 |
511421331141 has 32 divisors (see below), whose sum is σ = 866016569600. Its totient is φ = 292200929280.
The previous prime is 511421331139. The next prime is 511421331167. The reversal of 511421331141 is 141133124115.
511421331141 is a `hidden beast` number, since 51 + 142 + 1 + 331 + 141 = 666.
It is not a de Polignac number, because 511421331141 - 21 = 511421331139 is a prime.
It is a Harshad number since it is a multiple of its sum of digits (27).
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (511421331181) by changing a digit.
It is a polite number, since it can be written in 31 ways as a sum of consecutive naturals, for example, 1234761 + ... + 1596081.
It is an arithmetic number, because the mean of its divisors is an integer number (27063017800).
Almost surely, 2511421331141 is an apocalyptic number.
It is an amenable number.
511421331141 is a deficient number, since it is larger than the sum of its proper divisors (354595238459).
511421331141 is a wasteful number, since it uses less digits than its factorization.
511421331141 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 368826 (or 368820 counting only the distinct ones).
The product of its digits is 1440, while the sum is 27.
Adding to 511421331141 its reverse (141133124115), we get a palindrome (652554455256).
The spelling of 511421331141 in words is "five hundred eleven billion, four hundred twenty-one million, three hundred thirty-one thousand, one hundred forty-one".
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