Base | Representation |
---|---|
bin | 1011111011010001110100… |
… | …0001101000010111100111 |
3 | 1201102121000001120011220112 |
4 | 2332310131001220113213 |
5 | 3204320432130341242 |
6 | 43520012244523235 |
7 | 2522245624650503 |
oct | 276643501502747 |
9 | 51377001504815 |
10 | 13113022121447 |
11 | 41a622414a436 |
12 | 15794846b051b |
13 | 741725313761 |
14 | 33496072d303 |
15 | 17b176c42c82 |
hex | bed1d0685e7 |
13113022121447 has 2 divisors, whose sum is σ = 13113022121448. Its totient is φ = 13113022121446.
The previous prime is 13113022121267. The next prime is 13113022121477. The reversal of 13113022121447 is 74412122031131.
It is a strong prime.
It is an emirp because it is prime and its reverse (74412122031131) is a distict prime.
It is a cyclic number.
It is not a de Polignac number, because 13113022121447 - 218 = 13113021859303 is a prime.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (13113022121477) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6556511060723 + 6556511060724.
It is an arithmetic number, because the mean of its divisors is an integer number (6556511060724).
Almost surely, 213113022121447 is an apocalyptic number.
13113022121447 is a deficient number, since it is larger than the sum of its proper divisors (1).
13113022121447 is an equidigital number, since it uses as much as digits as its factorization.
13113022121447 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 8064, while the sum is 32.
Adding to 13113022121447 its reverse (74412122031131), we get a palindrome (87525144152578).
The spelling of 13113022121447 in words is "thirteen trillion, one hundred thirteen billion, twenty-two million, one hundred twenty-one thousand, four hundred forty-seven".
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