Base | Representation |
---|---|
bin | 11101110100010111111001… |
… | …011111011100111011010011 |
3 | 122012100002111212222010212101 |
4 | 131310113321133130323103 |
5 | 114142114140042023444 |
6 | 1142525555210541231 |
7 | 36423504322311244 |
oct | 3564277137347323 |
9 | 565302455863771 |
10 | 131142422220499 |
11 | 38871209516388 |
12 | 12860343a37817 |
13 | 58238a61ab681 |
14 | 24554817d8ccb |
15 | 10264ae4e63d4 |
hex | 7745f97dced3 |
131142422220499 has 2 divisors, whose sum is σ = 131142422220500. Its totient is φ = 131142422220498.
The previous prime is 131142422220479. The next prime is 131142422220517. The reversal of 131142422220499 is 994022224241131.
It is a strong prime.
It is an emirp because it is prime and its reverse (994022224241131) is a distict prime.
It is a cyclic number.
It is not a de Polignac number, because 131142422220499 - 215 = 131142422187731 is a prime.
It is not a weakly prime, because it can be changed into another prime (131142422220409) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (31) of ones.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 65571211110249 + 65571211110250.
It is an arithmetic number, because the mean of its divisors is an integer number (65571211110250).
Almost surely, 2131142422220499 is an apocalyptic number.
131142422220499 is a deficient number, since it is larger than the sum of its proper divisors (1).
131142422220499 is an equidigital number, since it uses as much as digits as its factorization.
131142422220499 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 497664, while the sum is 46.
The spelling of 131142422220499 in words is "one hundred thirty-one trillion, one hundred forty-two billion, four hundred twenty-two million, two hundred twenty thousand, four hundred ninety-nine".
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