Base | Representation |
---|---|
bin | 11110101111111111110… |
… | …110100010111100001011 |
3 | 21111000100001202222010002 |
4 | 132233333312202330023 |
5 | 234110133041300212 |
6 | 4254430553513215 |
7 | 305445055543235 |
oct | 36577766427413 |
9 | 7430301688102 |
10 | 2113121431307 |
11 | 745195a60497 |
12 | 2a165411280b |
13 | 124360651043 |
14 | 743c0449255 |
15 | 39e79153ac2 |
hex | 1ebffda2f0b |
2113121431307 has 2 divisors, whose sum is σ = 2113121431308. Its totient is φ = 2113121431306.
The previous prime is 2113121431301. The next prime is 2113121431313. The reversal of 2113121431307 is 7031341213112.
It is a balanced prime because it is at equal distance from previous prime (2113121431301) and next prime (2113121431313).
It is a cyclic number.
It is not a de Polignac number, because 2113121431307 - 28 = 2113121431051 is a prime.
It is a self number, because there is not a number n which added to its sum of digits gives 2113121431307.
It is not a weakly prime, because it can be changed into another prime (2113121431301) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 1056560715653 + 1056560715654.
It is an arithmetic number, because the mean of its divisors is an integer number (1056560715654).
Almost surely, 22113121431307 is an apocalyptic number.
2113121431307 is a deficient number, since it is larger than the sum of its proper divisors (1).
2113121431307 is an equidigital number, since it uses as much as digits as its factorization.
2113121431307 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 3024, while the sum is 29.
Adding to 2113121431307 its reverse (7031341213112), we get a palindrome (9144462644419).
The spelling of 2113121431307 in words is "two trillion, one hundred thirteen billion, one hundred twenty-one million, four hundred thirty-one thousand, three hundred seven".
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