Base | Representation |
---|---|
bin | 10100011111010100… |
… | …11111011100001011 |
3 | 1001101121211211120202 |
4 | 22033222133130023 |
5 | 140012023342212 |
6 | 5015312442415 |
7 | 536411200364 |
oct | 121752373413 |
9 | 31347754522 |
10 | 11000215307 |
11 | 4735375634 |
12 | 216bb4940b |
13 | 1063ca1653 |
14 | 764d28b6b |
15 | 445ad30c2 |
hex | 28fa9f70b |
11000215307 has 2 divisors, whose sum is σ = 11000215308. Its totient is φ = 11000215306.
The previous prime is 11000215283. The next prime is 11000215337. The reversal of 11000215307 is 70351200011.
It is a weak prime.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-11000215307 is a prime.
It is not a weakly prime, because it can be changed into another prime (11000215337) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 5500107653 + 5500107654.
It is an arithmetic number, because the mean of its divisors is an integer number (5500107654).
Almost surely, 211000215307 is an apocalyptic number.
11000215307 is a deficient number, since it is larger than the sum of its proper divisors (1).
11000215307 is an equidigital number, since it uses as much as digits as its factorization.
11000215307 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 210, while the sum is 20.
Adding to 11000215307 its reverse (70351200011), we get a palindrome (81351415318).
The spelling of 11000215307 in words is "eleven billion, two hundred fifteen thousand, three hundred seven".
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