Base | Representation |
---|---|
bin | 1011111011110000010100… |
… | …0101000110001010000111 |
3 | 1201110101010201002022202111 |
4 | 2332330011011012022013 |
5 | 3204434214314430203 |
6 | 43523444550121451 |
7 | 2522655552266233 |
oct | 276740505061207 |
9 | 51411121068674 |
10 | 13121210311303 |
11 | 41a9746173696 |
12 | 157ab8a950287 |
13 | 74242b837403 |
14 | 3350d9dc35c3 |
15 | 17b4a5a1326d |
hex | bef05146287 |
13121210311303 has 4 divisors (see below), whose sum is σ = 13121815471128. Its totient is φ = 13120605151480.
The previous prime is 13121210311243. The next prime is 13121210311327. The reversal of 13121210311303 is 30311301212131.
It is a semiprime because it is the product of two primes.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-13121210311303 is a prime.
It is a Duffinian number.
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (13121210311393) by changing a digit.
It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 302547388 + ... + 302590753.
It is an arithmetic number, because the mean of its divisors is an integer number (3280453867782).
Almost surely, 213121210311303 is an apocalyptic number.
13121210311303 is a deficient number, since it is larger than the sum of its proper divisors (605159825).
13121210311303 is an equidigital number, since it uses as much as digits as its factorization.
13121210311303 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 605159824.
The product of its (nonzero) digits is 324, while the sum is 22.
Adding to 13121210311303 its reverse (30311301212131), we get a palindrome (43432511523434).
The spelling of 13121210311303 in words is "thirteen trillion, one hundred twenty-one billion, two hundred ten million, three hundred eleven thousand, three hundred three".
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