Base | Representation |
---|---|
bin | 1011111011010010100001… |
… | …0010001100111011000011 |
3 | 1201102121111012222221221112 |
4 | 2332310220102030323003 |
5 | 3204321324004433401 |
6 | 43520043113313535 |
7 | 2522253410346233 |
oct | 276645022147303 |
9 | 51377435887845 |
10 | 13113211014851 |
11 | 41a6310836892 |
12 | 1579517a058ab |
13 | 7417554ac5bc |
14 | 33497b85dcc3 |
15 | 17b1886063bb |
hex | bed2848cec3 |
13113211014851 has 2 divisors, whose sum is σ = 13113211014852. Its totient is φ = 13113211014850.
The previous prime is 13113211014847. The next prime is 13113211014881. The reversal of 13113211014851 is 15841011231131.
13113211014851 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 13113211014851 - 22 = 13113211014847 is a prime.
It is not a weakly prime, because it can be changed into another prime (13113211014881) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6556605507425 + 6556605507426.
It is an arithmetic number, because the mean of its divisors is an integer number (6556605507426).
Almost surely, 213113211014851 is an apocalyptic number.
13113211014851 is a deficient number, since it is larger than the sum of its proper divisors (1).
13113211014851 is an equidigital number, since it uses as much as digits as its factorization.
13113211014851 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 2880, while the sum is 32.
Adding to 13113211014851 its reverse (15841011231131), we get a palindrome (28954222245982).
The spelling of 13113211014851 in words is "thirteen trillion, one hundred thirteen billion, two hundred eleven million, fourteen thousand, eight hundred fifty-one".
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