Base | Representation |
---|---|
bin | 1100010101000000011101… |
… | …0111111100000011111011 |
3 | 1202222211221002010110000112 |
4 | 3011100013113330003323 |
5 | 3234041210332031021 |
6 | 44455033304454535 |
7 | 2566214353126661 |
oct | 305200727740373 |
9 | 52884832113015 |
10 | 13555040502011 |
11 | 4356729a90a88 |
12 | 162b08377a44b |
13 | 774308c15978 |
14 | 34c0d3180c31 |
15 | 1878e7416d5b |
hex | c54075fc0fb |
13555040502011 has 2 divisors, whose sum is σ = 13555040502012. Its totient is φ = 13555040502010.
The previous prime is 13555040501993. The next prime is 13555040502109. The reversal of 13555040502011 is 11020504055531.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 13555040502011 - 222 = 13555036307707 is a prime.
It is a super-2 number, since 2×135550405020112 (a number of 27 digits) contains 22 as substring.
It is not a weakly prime, because it can be changed into another prime (13555040502511) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (23) of ones.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6777520251005 + 6777520251006.
It is an arithmetic number, because the mean of its divisors is an integer number (6777520251006).
Almost surely, 213555040502011 is an apocalyptic number.
13555040502011 is a deficient number, since it is larger than the sum of its proper divisors (1).
13555040502011 is an equidigital number, since it uses as much as digits as its factorization.
13555040502011 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 15000, while the sum is 32.
Adding to 13555040502011 its reverse (11020504055531), we get a palindrome (24575544557542).
The spelling of 13555040502011 in words is "thirteen trillion, five hundred fifty-five billion, forty million, five hundred two thousand, eleven".
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