Base | Representation |
---|---|
bin | 11101111000100110000000… |
… | …001001101000100110010011 |
3 | 122020100211010000112221111211 |
4 | 131320212000021220212103 |
5 | 114211342121210443011 |
6 | 1143311121545120551 |
7 | 36453452330615425 |
oct | 3570460011504623 |
9 | 566324100487454 |
10 | 131432444234131 |
11 | 38973206373857 |
12 | 128a85a3789157 |
13 | 5845045a59921 |
14 | 24655156a8815 |
15 | 102dcd5be4b21 |
hex | 778980268993 |
131432444234131 has 2 divisors, whose sum is σ = 131432444234132. Its totient is φ = 131432444234130.
The previous prime is 131432444234051. The next prime is 131432444234177.
131432444234131 is nontrivially palindromic in base 10.
It is a strong prime.
It is a palprime.
It is a cyclic number.
It is not a de Polignac number, because 131432444234131 - 213 = 131432444225939 is a prime.
It is a super-2 number, since 2×1314324442341312 (a number of 29 digits) contains 22 as substring.
It is not a weakly prime, because it can be changed into another prime (131432444234431) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 65716222117065 + 65716222117066.
It is an arithmetic number, because the mean of its divisors is an integer number (65716222117066).
Almost surely, 2131432444234131 is an apocalyptic number.
131432444234131 is a deficient number, since it is larger than the sum of its proper divisors (1).
131432444234131 is an equidigital number, since it uses as much as digits as its factorization.
131432444234131 is an evil number, because the sum of its binary digits is even.
The product of its digits is 331776, while the sum is 40.
The spelling of 131432444234131 in words is "one hundred thirty-one trillion, four hundred thirty-two billion, four hundred forty-four million, two hundred thirty-four thousand, one hundred thirty-one".
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