Base | Representation |
---|---|
bin | 11101100101110011100001… |
… | …111010001110010101000111 |
3 | 122001210101102220101122100021 |
4 | 131211303201322032111013 |
5 | 114024213340044033242 |
6 | 1140442030522200011 |
7 | 36261253500141655 |
oct | 3545634172162507 |
9 | 561711386348307 |
10 | 130141299205447 |
11 | 38515684696682 |
12 | 1271a309953007 |
13 | 578036c3645aa |
14 | 241cc30181dd5 |
15 | 100a41939eb67 |
hex | 765ce1e8e547 |
130141299205447 has 2 divisors, whose sum is σ = 130141299205448. Its totient is φ = 130141299205446.
The previous prime is 130141299205391. The next prime is 130141299205471. The reversal of 130141299205447 is 744502992141031.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 130141299205447 - 223 = 130141290816839 is a prime.
It is a junction number, because it is equal to n+sod(n) for n = 130141299205394 and 130141299205403.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (130141299205847) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 65070649602723 + 65070649602724.
It is an arithmetic number, because the mean of its divisors is an integer number (65070649602724).
Almost surely, 2130141299205447 is an apocalyptic number.
130141299205447 is a deficient number, since it is larger than the sum of its proper divisors (1).
130141299205447 is an equidigital number, since it uses as much as digits as its factorization.
130141299205447 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 2177280, while the sum is 52.
The spelling of 130141299205447 in words is "one hundred thirty trillion, one hundred forty-one billion, two hundred ninety-nine million, two hundred five thousand, four hundred forty-seven".
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