Base | Representation |
---|---|
bin | 1010000111010110001100… |
… | …0000010000111111110111 |
3 | 1110101012002200222112112212 |
4 | 2201311203000100333313 |
5 | 2424202441310441401 |
6 | 35353022531204035 |
7 | 2225326646451245 |
oct | 241654300207767 |
9 | 43335080875485 |
10 | 11121331343351 |
11 | 35a8591679a99 |
12 | 12b747b44661b |
13 | 62897739b948 |
14 | 2a63bd63b795 |
15 | 144458491dbb |
hex | a1d63010ff7 |
11121331343351 has 2 divisors, whose sum is σ = 11121331343352. Its totient is φ = 11121331343350.
The previous prime is 11121331343281. The next prime is 11121331343357. The reversal of 11121331343351 is 15334313312111.
11121331343351 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 11121331343351 - 230 = 11120257601527 is a prime.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (11121331343357) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 5560665671675 + 5560665671676.
It is an arithmetic number, because the mean of its divisors is an integer number (5560665671676).
Almost surely, 211121331343351 is an apocalyptic number.
11121331343351 is a deficient number, since it is larger than the sum of its proper divisors (1).
11121331343351 is an equidigital number, since it uses as much as digits as its factorization.
11121331343351 is an evil number, because the sum of its binary digits is even.
The product of its digits is 9720, while the sum is 32.
Adding to 11121331343351 its reverse (15334313312111), we get a palindrome (26455644655462).
The spelling of 11121331343351 in words is "eleven trillion, one hundred twenty-one billion, three hundred thirty-one million, three hundred forty-three thousand, three hundred fifty-one".
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