Base | Representation |
---|---|
bin | 10100111000101110… |
… | …11101101000110001 |
3 | 1001221110212020100022 |
4 | 22130113131220301 |
5 | 140431103042433 |
6 | 5052404333225 |
7 | 544606512044 |
oct | 123427355061 |
9 | 31843766308 |
10 | 11213330993 |
11 | 48346a65a7 |
12 | 220b3a4215 |
13 | 109919b6c7 |
14 | 785362c5b |
15 | 45967d698 |
hex | 29c5dda31 |
11213330993 has 2 divisors, whose sum is σ = 11213330994. Its totient is φ = 11213330992.
The previous prime is 11213330981. The next prime is 11213331001. The reversal of 11213330993 is 39903331211.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 7626878224 + 3586452769 = 87332^2 + 59887^2 .
It is an emirp because it is prime and its reverse (39903331211) is a distict prime.
It is a cyclic number.
It is not a de Polignac number, because 11213330993 - 222 = 11209136689 is a prime.
It is a super-3 number, since 3×112133309933 (a number of 31 digits) contains 333 as substring.
It is not a weakly prime, because it can be changed into another prime (11413330993) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 5606665496 + 5606665497.
It is an arithmetic number, because the mean of its divisors is an integer number (5606665497).
Almost surely, 211213330993 is an apocalyptic number.
It is an amenable number.
11213330993 is a deficient number, since it is larger than the sum of its proper divisors (1).
11213330993 is an equidigital number, since it uses as much as digits as its factorization.
11213330993 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 13122, while the sum is 35.
The spelling of 11213330993 in words is "eleven billion, two hundred thirteen million, three hundred thirty thousand, nine hundred ninety-three".
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