Base | Representation |
---|---|
bin | 1010010100111001010101… |
… | …0110010010011011001011 |
3 | 1111012102221001012101221122 |
4 | 2211032111112102123023 |
5 | 2442011201134224042 |
6 | 40052000425425455 |
7 | 2251210200256625 |
oct | 245162526223313 |
9 | 44172831171848 |
10 | 11354104211147 |
11 | 368828090a8a5 |
12 | 133460263b28b |
13 | 6448c23a0589 |
14 | 2b3782127c15 |
15 | 14a52dbba7d2 |
hex | a53955926cb |
11354104211147 has 2 divisors, whose sum is σ = 11354104211148. Its totient is φ = 11354104211146.
The previous prime is 11354104211083. The next prime is 11354104211227. The reversal of 11354104211147 is 74111240145311.
11354104211147 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 11354104211147 - 26 = 11354104211083 is a prime.
It is not a weakly prime, because it can be changed into another prime (11354104211347) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 5677052105573 + 5677052105574.
It is an arithmetic number, because the mean of its divisors is an integer number (5677052105574).
Almost surely, 211354104211147 is an apocalyptic number.
11354104211147 is a deficient number, since it is larger than the sum of its proper divisors (1).
11354104211147 is an equidigital number, since it uses as much as digits as its factorization.
11354104211147 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 13440, while the sum is 35.
Adding to 11354104211147 its reverse (74111240145311), we get a palindrome (85465344356458).
The spelling of 11354104211147 in words is "eleven trillion, three hundred fifty-four billion, one hundred four million, two hundred eleven thousand, one hundred forty-seven".
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