Base | Representation |
---|---|
bin | 111111000001111100001… |
… | …110100111100010110011 |
3 | 120100002010111102002221201 |
4 | 333001330032213202303 |
5 | 1031431212234204024 |
6 | 13113453423032031 |
7 | 624635260036123 |
oct | 77017416474263 |
9 | 16302114362851 |
10 | 4331411241139 |
11 | 141aa397a9294 |
12 | 59b558a3b617 |
13 | 2555b14a5658 |
14 | 10d8da0c6c83 |
15 | 77a0b108744 |
hex | 3f07c3a78b3 |
4331411241139 has 2 divisors, whose sum is σ = 4331411241140. Its totient is φ = 4331411241138.
The previous prime is 4331411241121. The next prime is 4331411241161. The reversal of 4331411241139 is 9311421141334.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 4331411241139 - 211 = 4331411239091 is a prime.
It is a super-2 number, since 2×43314112411392 (a number of 26 digits) contains 22 as substring.
It is a junction number, because it is equal to n+sod(n) for n = 4331411241098 and 4331411241107.
It is not a weakly prime, because it can be changed into another prime (4331411242139) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 2165705620569 + 2165705620570.
It is an arithmetic number, because the mean of its divisors is an integer number (2165705620570).
Almost surely, 24331411241139 is an apocalyptic number.
4331411241139 is a deficient number, since it is larger than the sum of its proper divisors (1).
4331411241139 is an equidigital number, since it uses as much as digits as its factorization.
4331411241139 is an evil number, because the sum of its binary digits is even.
The product of its digits is 31104, while the sum is 37.
The spelling of 4331411241139 in words is "four trillion, three hundred thirty-one billion, four hundred eleven million, two hundred forty-one thousand, one hundred thirty-nine".
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