Base | Representation |
---|---|
bin | 11110000010101111101011… |
… | …101000100101100010110011 |
3 | 122022211111201010101221010012 |
4 | 132002233223220211202303 |
5 | 114304304301221242042 |
6 | 1145003410540325135 |
7 | 36555036241605044 |
oct | 3602575350454263 |
9 | 568744633357105 |
10 | 132130032212147 |
11 | 39112039619652 |
12 | 1299b8285437ab |
13 | 5895a6858bc08 |
14 | 248b1b05008cb |
15 | 104201314de82 |
hex | 782beba258b3 |
132130032212147 has 2 divisors, whose sum is σ = 132130032212148. Its totient is φ = 132130032212146.
The previous prime is 132130032212117. The next prime is 132130032212161. The reversal of 132130032212147 is 741212230031231.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 132130032212147 - 230 = 132128958470323 is a prime.
It is not a weakly prime, because it can be changed into another prime (132130032212117) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 66065016106073 + 66065016106074.
It is an arithmetic number, because the mean of its divisors is an integer number (66065016106074).
Almost surely, 2132130032212147 is an apocalyptic number.
132130032212147 is a deficient number, since it is larger than the sum of its proper divisors (1).
132130032212147 is an equidigital number, since it uses as much as digits as its factorization.
132130032212147 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 12096, while the sum is 32.
Adding to 132130032212147 its reverse (741212230031231), we get a palindrome (873342262243378).
The spelling of 132130032212147 in words is "one hundred thirty-two trillion, one hundred thirty billion, thirty-two million, two hundred twelve thousand, one hundred forty-seven".
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