Base | Representation |
---|---|
bin | 10011011000101001110… |
… | …010110101110011100111 |
3 | 11201100111012102020012021 |
4 | 103120221302311303213 |
5 | 133311211102231312 |
6 | 2455551022004011 |
7 | 165146461065604 |
oct | 23305162656347 |
9 | 4640435366167 |
10 | 1332141055207 |
11 | 473a59995736 |
12 | 196217416607 |
13 | 9880ac232c1 |
14 | 486940c96ab |
15 | 249ba9c5c07 |
hex | 13629cb5ce7 |
1332141055207 has 2 divisors, whose sum is σ = 1332141055208. Its totient is φ = 1332141055206.
The previous prime is 1332141055187. The next prime is 1332141055247. The reversal of 1332141055207 is 7025501412331.
It is a weak prime.
It is an emirp because it is prime and its reverse (7025501412331) is a distict prime.
It is a cyclic number.
It is not a de Polignac number, because 1332141055207 - 215 = 1332141022439 is a prime.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (1332141055247) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (23) of ones.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 666070527603 + 666070527604.
It is an arithmetic number, because the mean of its divisors is an integer number (666070527604).
Almost surely, 21332141055207 is an apocalyptic number.
1332141055207 is a deficient number, since it is larger than the sum of its proper divisors (1).
1332141055207 is an equidigital number, since it uses as much as digits as its factorization.
1332141055207 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 25200, while the sum is 34.
Adding to 1332141055207 its reverse (7025501412331), we get a palindrome (8357642467538).
The spelling of 1332141055207 in words is "one trillion, three hundred thirty-two billion, one hundred forty-one million, fifty-five thousand, two hundred seven".
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