Base | Representation |
---|---|
bin | 11101110011001010100010… |
… | …110010101011010000111111 |
3 | 122012001011010100212221210122 |
4 | 131303022202302223100333 |
5 | 114134234033443413201 |
6 | 1142423501301125155 |
7 | 36414504124652642 |
oct | 3563124262532077 |
9 | 565034110787718 |
10 | 131059363263551 |
11 | 38839a66940708 |
12 | 128482236b97bb |
13 | 5818aca3bc078 |
14 | 24514426aba59 |
15 | 102424c6d9e1b |
hex | 7732a2cab43f |
131059363263551 has 2 divisors, whose sum is σ = 131059363263552. Its totient is φ = 131059363263550.
The previous prime is 131059363263547. The next prime is 131059363263559. The reversal of 131059363263551 is 155362363950131.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 131059363263551 - 22 = 131059363263547 is a prime.
It is a junction number, because it is equal to n+sod(n) for n = 131059363263493 and 131059363263502.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (131059363263559) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 65529681631775 + 65529681631776.
It is an arithmetic number, because the mean of its divisors is an integer number (65529681631776).
Almost surely, 2131059363263551 is an apocalyptic number.
131059363263551 is a deficient number, since it is larger than the sum of its proper divisors (1).
131059363263551 is an equidigital number, since it uses as much as digits as its factorization.
131059363263551 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 6561000, while the sum is 53.
The spelling of 131059363263551 in words is "one hundred thirty-one trillion, fifty-nine billion, three hundred sixty-three million, two hundred sixty-three thousand, five hundred fifty-one".
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