Base | Representation |
---|---|
bin | 1011111011001010101100… |
… | …0111110000000101000011 |
3 | 1201102102001212200102020011 |
4 | 2332302223013300011003 |
5 | 3204303023421421032 |
6 | 43515054501325351 |
7 | 2522150361513313 |
oct | 276625307600503 |
9 | 51372055612204 |
10 | 13111111123267 |
11 | 41a54334721a3 |
12 | 1579030706857 |
13 | 7414ac42bbc1 |
14 | 33481ca01443 |
15 | 17b0b40b1247 |
hex | becab1f0143 |
13111111123267 has 2 divisors, whose sum is σ = 13111111123268. Its totient is φ = 13111111123266.
The previous prime is 13111111123259. The next prime is 13111111123333. The reversal of 13111111123267 is 76232111111131.
13111111123267 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 13111111123267 - 23 = 13111111123259 is a prime.
It is not a weakly prime, because it can be changed into another prime (13111111123367) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6555555561633 + 6555555561634.
It is an arithmetic number, because the mean of its divisors is an integer number (6555555561634).
Almost surely, 213111111123267 is an apocalyptic number.
13111111123267 is a deficient number, since it is larger than the sum of its proper divisors (1).
13111111123267 is an equidigital number, since it uses as much as digits as its factorization.
13111111123267 is an evil number, because the sum of its binary digits is even.
The product of its digits is 1512, while the sum is 31.
Adding to 13111111123267 its reverse (76232111111131), we get a palindrome (89343222234398).
The spelling of 13111111123267 in words is "thirteen trillion, one hundred eleven billion, one hundred eleven million, one hundred twenty-three thousand, two hundred sixty-seven".
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