Base | Representation |
---|---|
bin | 10100101111000010… |
… | …11010011001111101 |
3 | 1001201210220220001121 |
4 | 22113201122121331 |
5 | 140244301144432 |
6 | 5040342310541 |
7 | 542601614233 |
oct | 122741323175 |
9 | 31653826047 |
10 | 11132053117 |
11 | 47a28322a9 |
12 | 21a8128451 |
13 | 10853b17a3 |
14 | 778646953 |
15 | 452476197 |
hex | 29785a67d |
11132053117 has 2 divisors, whose sum is σ = 11132053118. Its totient is φ = 11132053116.
The previous prime is 11132053103. The next prime is 11132053157. The reversal of 11132053117 is 71135023111.
It is a weak prime.
It can be written as a sum of positive squares in only one way, i.e., 7761786201 + 3370266916 = 88101^2 + 58054^2 .
It is a cyclic number.
It is not a de Polignac number, because 11132053117 - 211 = 11132051069 is a prime.
It is a junction number, because it is equal to n+sod(n) for n = 11132053091 and 11132053100.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (11132053157) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (19) of ones.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 5566026558 + 5566026559.
It is an arithmetic number, because the mean of its divisors is an integer number (5566026559).
Almost surely, 211132053117 is an apocalyptic number.
It is an amenable number.
11132053117 is a deficient number, since it is larger than the sum of its proper divisors (1).
11132053117 is an equidigital number, since it uses as much as digits as its factorization.
11132053117 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 630, while the sum is 25.
Adding to 11132053117 its reverse (71135023111), we get a palindrome (82267076228).
The spelling of 11132053117 in words is "eleven billion, one hundred thirty-two million, fifty-three thousand, one hundred seventeen".
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