Base | Representation |
---|---|
bin | 1100110010001001000001… |
… | …1000111101010111000011 |
3 | 1211202200211211022110010212 |
4 | 3030202100120331113003 |
5 | 3320241234144020103 |
6 | 45521011045144335 |
7 | 2650323524050301 |
oct | 314422030752703 |
9 | 54680754273125 |
10 | 14055555454403 |
11 | 4529a22698871 |
12 | 16b00890bb0ab |
13 | 7ac5829525a8 |
14 | 36841492d271 |
15 | 19593d3687d8 |
hex | cc89063d5c3 |
14055555454403 has 2 divisors, whose sum is σ = 14055555454404. Its totient is φ = 14055555454402.
The previous prime is 14055555454357. The next prime is 14055555454427. The reversal of 14055555454403 is 30445455555041.
It is a strong prime.
It is an emirp because it is prime and its reverse (30445455555041) is a distict prime.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-14055555454403 is a prime.
It is not a weakly prime, because it can be changed into another prime (14055555454433) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 7027777727201 + 7027777727202.
It is an arithmetic number, because the mean of its divisors is an integer number (7027777727202).
Almost surely, 214055555454403 is an apocalyptic number.
14055555454403 is a deficient number, since it is larger than the sum of its proper divisors (1).
14055555454403 is an equidigital number, since it uses as much as digits as its factorization.
14055555454403 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 12000000, while the sum is 50.
The spelling of 14055555454403 in words is "fourteen trillion, fifty-five billion, five hundred fifty-five million, four hundred fifty-four thousand, four hundred three".
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