Base | Representation |
---|---|
bin | 11110111010010101100… |
… | …111100100001010001011 |
3 | 21112001222211020222010121 |
4 | 132322111213210022023 |
5 | 234300402322134011 |
6 | 4303504404524111 |
7 | 306320152155622 |
oct | 36722547441213 |
9 | 7461884228117 |
10 | 2124224021131 |
11 | 749973099563 |
12 | 2a38323b0637 |
13 | 12540c8b7c14 |
14 | 74b54bc26b9 |
15 | 3a3c8c09e71 |
hex | 1ee959e428b |
2124224021131 has 2 divisors, whose sum is σ = 2124224021132. Its totient is φ = 2124224021130.
The previous prime is 2124224021123. The next prime is 2124224021177. The reversal of 2124224021131 is 1311204224212.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 2124224021131 - 23 = 2124224021123 is a prime.
It is a super-4 number, since 4×21242240211314 (a number of 50 digits) contains 4444 as substring. Note that it is a super-d number also for d = 2.
It is a junction number, because it is equal to n+sod(n) for n = 2124224021096 and 2124224021105.
It is not a weakly prime, because it can be changed into another prime (2124224031131) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 1062112010565 + 1062112010566.
It is an arithmetic number, because the mean of its divisors is an integer number (1062112010566).
Almost surely, 22124224021131 is an apocalyptic number.
2124224021131 is a deficient number, since it is larger than the sum of its proper divisors (1).
2124224021131 is an equidigital number, since it uses as much as digits as its factorization.
2124224021131 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 1536, while the sum is 25.
Adding to 2124224021131 its reverse (1311204224212), we get a palindrome (3435428245343).
The spelling of 2124224021131 in words is "two trillion, one hundred twenty-four billion, two hundred twenty-four million, twenty-one thousand, one hundred thirty-one".
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