Base | Representation |
---|---|
bin | 10100011111010100… |
… | …00011010100100011 |
3 | 1001101121121222120222 |
4 | 22033222003110203 |
5 | 140012011201011 |
6 | 5015310153255 |
7 | 536410210526 |
oct | 121752032443 |
9 | 31347558528 |
10 | 11000100131 |
11 | 47352a7049 |
12 | 216bab282b |
13 | 1063c610b7 |
14 | 764cd8bbd |
15 | 445aadddb |
hex | 28fa83523 |
11000100131 has 2 divisors, whose sum is σ = 11000100132. Its totient is φ = 11000100130.
The previous prime is 11000100107. The next prime is 11000100143. The reversal of 11000100131 is 13100100011.
It is a strong prime.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-11000100131 is a prime.
It is a super-2 number, since 2×110001001312 (a number of 21 digits) contains 22 as substring.
It is not a weakly prime, because it can be changed into another prime (11000100031) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 5500050065 + 5500050066.
It is an arithmetic number, because the mean of its divisors is an integer number (5500050066).
Almost surely, 211000100131 is an apocalyptic number.
11000100131 is a deficient number, since it is larger than the sum of its proper divisors (1).
11000100131 is an equidigital number, since it uses as much as digits as its factorization.
11000100131 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 3, while the sum is 8.
Adding to 11000100131 its reverse (13100100011), we get a palindrome (24100200142).
The spelling of 11000100131 in words is "eleven billion, one hundred thousand, one hundred thirty-one".
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