Base | Representation |
---|---|
bin | 11101111110010011101101… |
… | …110001011111110010011111 |
3 | 122021202100200122111020211101 |
4 | 131332103231301133302133 |
5 | 114234310324102034122 |
6 | 1144211343201533531 |
7 | 36524023332636421 |
oct | 3576235561376237 |
9 | 567670618436741 |
10 | 131825125424287 |
11 | 390047a369aa79 |
12 | 129507139442a7 |
13 | 5873096009749 |
14 | 247a527842811 |
15 | 1039119cec027 |
hex | 77e4edc5fc9f |
131825125424287 has 2 divisors, whose sum is σ = 131825125424288. Its totient is φ = 131825125424286.
The previous prime is 131825125424281. The next prime is 131825125424351. The reversal of 131825125424287 is 782424521528131.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 131825125424287 - 243 = 123029032402079 is a prime.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (131825125424281) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 65912562712143 + 65912562712144.
It is an arithmetic number, because the mean of its divisors is an integer number (65912562712144).
Almost surely, 2131825125424287 is an apocalyptic number.
131825125424287 is a deficient number, since it is larger than the sum of its proper divisors (1).
131825125424287 is an equidigital number, since it uses as much as digits as its factorization.
131825125424287 is an evil number, because the sum of its binary digits is even.
The product of its digits is 8601600, while the sum is 55.
The spelling of 131825125424287 in words is "one hundred thirty-one trillion, eight hundred twenty-five billion, one hundred twenty-five million, four hundred twenty-four thousand, two hundred eighty-seven".
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