Base | Representation |
---|---|
bin | 11101010010111100000… |
… | …101000100100010010111 |
3 | 21010110102101020110111102 |
4 | 131102330011010202113 |
5 | 230441013213443224 |
6 | 4140503521500315 |
7 | 265306655563355 |
oct | 35227405044227 |
9 | 7113371213442 |
10 | 2013200140439 |
11 | 70687aaa4764 |
12 | 2862087a169b |
13 | 117ac823c267 |
14 | 6d621a367d5 |
15 | 3757bcb54ae |
hex | 1d4bc144897 |
2013200140439 has 2 divisors, whose sum is σ = 2013200140440. Its totient is φ = 2013200140438.
The previous prime is 2013200140379. The next prime is 2013200140463. The reversal of 2013200140439 is 9340410023102.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 2013200140439 - 28 = 2013200140183 is a prime.
It is a super-2 number, since 2×20132001404392 (a number of 25 digits) contains 22 as substring.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (2013200140139) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (19) of ones.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 1006600070219 + 1006600070220.
It is an arithmetic number, because the mean of its divisors is an integer number (1006600070220).
Almost surely, 22013200140439 is an apocalyptic number.
2013200140439 is a deficient number, since it is larger than the sum of its proper divisors (1).
2013200140439 is an equidigital number, since it uses as much as digits as its factorization.
2013200140439 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 5184, while the sum is 29.
The spelling of 2013200140439 in words is "two trillion, thirteen billion, two hundred million, one hundred forty thousand, four hundred thirty-nine".
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