Base | Representation |
---|---|
bin | 10000001010111001011… |
… | …101001100010110111111 |
3 | 10221020020100022012012202 |
4 | 100022321131030112333 |
5 | 121201231012001002 |
6 | 2210252331022115 |
7 | 143165615441006 |
oct | 20127135142677 |
9 | 3836210265182 |
10 | 1111213000127 |
11 | 399298850a4a |
12 | 15b43ab9a93b |
13 | 80a2cb8c1a9 |
14 | 3bad6781c3d |
15 | 1dd8a0b6502 |
hex | 102b974c5bf |
1111213000127 has 2 divisors, whose sum is σ = 1111213000128. Its totient is φ = 1111213000126.
The previous prime is 1111213000091. The next prime is 1111213000159. The reversal of 1111213000127 is 7210003121111.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 1111213000127 - 212 = 1111212996031 is a prime.
It is a super-2 number, since 2×11112130001272 (a number of 25 digits) contains 22 as substring.
It is a junction number, because it is equal to n+sod(n) for n = 1111213000099 and 1111213000108.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (1111213080127) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 555606500063 + 555606500064.
It is an arithmetic number, because the mean of its divisors is an integer number (555606500064).
Almost surely, 21111213000127 is an apocalyptic number.
1111213000127 is a deficient number, since it is larger than the sum of its proper divisors (1).
1111213000127 is an equidigital number, since it uses as much as digits as its factorization.
1111213000127 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 84, while the sum is 20.
Adding to 1111213000127 its reverse (7210003121111), we get a palindrome (8321216121238).
The spelling of 1111213000127 in words is "one trillion, one hundred eleven billion, two hundred thirteen million, one hundred twenty-seven", and thus it is an aban number.
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