Base | Representation |
---|---|
bin | 11100111111101000… |
… | …110000110100111001 |
3 | 2222100200010212112200 |
4 | 130333220300310321 |
5 | 1002224344441213 |
6 | 22145123500413 |
7 | 2151330134031 |
oct | 347750606471 |
9 | 88320125480 |
10 | 31132421433 |
11 | 12226493013 |
12 | 604a227709 |
13 | 2c21b87c59 |
14 | 17149c68c1 |
15 | c23266273 |
hex | 73fa30d39 |
31132421433 has 12 divisors (see below), whose sum is σ = 45170711040. Its totient is φ = 20661874776.
The previous prime is 31132421423. The next prime is 31132421447. The reversal of 31132421433 is 33412423113.
It is a happy number.
31132421433 is a `hidden beast` number, since 311 + 324 + 21 + 4 + 3 + 3 = 666.
It is a de Polignac number, because none of the positive numbers 2k-31132421433 is a prime.
It is a junction number, because it is equal to n+sod(n) for n = 31132421397 and 31132421406.
It is not an unprimeable number, because it can be changed into a prime (31132421423) by changing a digit.
It is a polite number, since it can be written in 11 ways as a sum of consecutive naturals, for example, 7753953 + ... + 7757966.
It is an arithmetic number, because the mean of its divisors is an integer number (3764225920).
Almost surely, 231132421433 is an apocalyptic number.
It is an amenable number.
31132421433 is a deficient number, since it is larger than the sum of its proper divisors (14038289607).
31132421433 is a wasteful number, since it uses less digits than its factorization.
31132421433 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 15512148 (or 15512145 counting only the distinct ones).
The product of its digits is 5184, while the sum is 27.
Adding to 31132421433 its reverse (33412423113), we get a palindrome (64544844546).
The spelling of 31132421433 in words is "thirty-one billion, one hundred thirty-two million, four hundred twenty-one thousand, four hundred thirty-three".
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