Base | Representation |
---|---|
bin | 11101010011000010101… |
… | …000001100011011110111 |
3 | 21010110201000222110202011 |
4 | 131103002220030123313 |
5 | 230441224330324002 |
6 | 4140522444425051 |
7 | 265312461506152 |
oct | 35230250143367 |
9 | 7113630873664 |
10 | 2013310011127 |
11 | 706927018188 |
12 | 286239548187 |
13 | 117b14c2a7bc |
14 | 6d632476c99 |
15 | 375867698d7 |
hex | 1d4c2a0c6f7 |
2013310011127 has 2 divisors, whose sum is σ = 2013310011128. Its totient is φ = 2013310011126.
The previous prime is 2013310011121. The next prime is 2013310011197. The reversal of 2013310011127 is 7211100133102.
It is a weak prime.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-2013310011127 is a prime.
It is a junction number, because it is equal to n+sod(n) for n = 2013310011098 and 2013310011107.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (2013310011121) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 1006655005563 + 1006655005564.
It is an arithmetic number, because the mean of its divisors is an integer number (1006655005564).
Almost surely, 22013310011127 is an apocalyptic number.
2013310011127 is a deficient number, since it is larger than the sum of its proper divisors (1).
2013310011127 is an equidigital number, since it uses as much as digits as its factorization.
2013310011127 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 252, while the sum is 22.
Adding to 2013310011127 its reverse (7211100133102), we get a palindrome (9224410144229).
The spelling of 2013310011127 in words is "two trillion, thirteen billion, three hundred ten million, eleven thousand, one hundred twenty-seven".
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