Base | Representation |
---|---|
bin | 11101101010000011010101… |
… | …000011001101100111010011 |
3 | 122002211021201212122210021121 |
4 | 131222003111003031213103 |
5 | 114044004033123443011 |
6 | 1141224054033503111 |
7 | 36321324555354361 |
oct | 3552032503154723 |
9 | 562737655583247 |
10 | 130433141234131 |
11 | 3861842592931a |
12 | 12766997111a97 |
13 | 57a1a3bcbcb28 |
14 | 242cdd7a58631 |
15 | 1012cea75be71 |
hex | 76a0d50cd9d3 |
130433141234131 has 2 divisors, whose sum is σ = 130433141234132. Its totient is φ = 130433141234130.
The previous prime is 130433141234129. The next prime is 130433141234177. The reversal of 130433141234131 is 131432141334031.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 130433141234131 - 21 = 130433141234129 is a prime.
It is a super-2 number, since 2×1304331412341312 (a number of 29 digits) contains 22 as substring.
Together with 130433141234129, it forms a pair of twin primes.
It is not a weakly prime, because it can be changed into another prime (130433141834131) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 65216570617065 + 65216570617066.
It is an arithmetic number, because the mean of its divisors is an integer number (65216570617066).
Almost surely, 2130433141234131 is an apocalyptic number.
130433141234131 is a deficient number, since it is larger than the sum of its proper divisors (1).
130433141234131 is an equidigital number, since it uses as much as digits as its factorization.
130433141234131 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 31104, while the sum is 34.
Adding to 130433141234131 its reverse (131432141334031), we get a palindrome (261865282568162).
The spelling of 130433141234131 in words is "one hundred thirty trillion, four hundred thirty-three billion, one hundred forty-one million, two hundred thirty-four thousand, one hundred thirty-one".
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