Base | Representation |
---|---|
bin | 11101110010011110101001… |
… | …000100110100000000010111 |
3 | 122011212122110011010100001111 |
4 | 131302132221010310000113 |
5 | 114133001013323302413 |
6 | 1142350103532035451 |
7 | 36411215021121115 |
oct | 3562365104640027 |
9 | 564778404110044 |
10 | 131012224040983 |
11 | 38820a76a9802a |
12 | 1283b068902b87 |
13 | 581452723936c |
14 | 244d04dca2ab5 |
15 | 1022dde087d3d |
hex | 7727a9134017 |
131012224040983 has 2 divisors, whose sum is σ = 131012224040984. Its totient is φ = 131012224040982.
The previous prime is 131012224040969. The next prime is 131012224041037. The reversal of 131012224040983 is 389040422210131.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 131012224040983 - 29 = 131012224040471 is a prime.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (131012224040083) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 65506112020491 + 65506112020492.
It is an arithmetic number, because the mean of its divisors is an integer number (65506112020492).
Almost surely, 2131012224040983 is an apocalyptic number.
131012224040983 is a deficient number, since it is larger than the sum of its proper divisors (1).
131012224040983 is an equidigital number, since it uses as much as digits as its factorization.
131012224040983 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 82944, while the sum is 40.
The spelling of 131012224040983 in words is "one hundred thirty-one trillion, twelve billion, two hundred twenty-four million, forty thousand, nine hundred eighty-three".
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