Base | Representation |
---|---|
bin | 11110000110110100011110… |
… | …010011001000111001011111 |
3 | 122100211021111101212121122122 |
4 | 132012310132103020321133 |
5 | 114323401243101231142 |
6 | 1145340201202533155 |
7 | 36614211413110052 |
oct | 3606643623107137 |
9 | 570737441777578 |
10 | 132410055102047 |
11 | 3920a8752003a8 |
12 | 12a25b576671bb |
13 | 58b62a36987c8 |
14 | 249a9763d8a99 |
15 | 1049451ab41d2 |
hex | 786d1e4c8e5f |
132410055102047 has 2 divisors, whose sum is σ = 132410055102048. Its totient is φ = 132410055102046.
The previous prime is 132410055102007. The next prime is 132410055102151. The reversal of 132410055102047 is 740201550014231.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 132410055102047 - 212 = 132410055097951 is a prime.
It is a junction number, because it is equal to n+sod(n) for n = 132410055101998 and 132410055102016.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (132410055102007) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 66205027551023 + 66205027551024.
It is an arithmetic number, because the mean of its divisors is an integer number (66205027551024).
Almost surely, 2132410055102047 is an apocalyptic number.
132410055102047 is a deficient number, since it is larger than the sum of its proper divisors (1).
132410055102047 is an equidigital number, since it uses as much as digits as its factorization.
132410055102047 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 33600, while the sum is 35.
The spelling of 132410055102047 in words is "one hundred thirty-two trillion, four hundred ten billion, fifty-five million, one hundred two thousand, forty-seven".
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