Base | Representation |
---|---|
bin | 1001001001011101001011… |
… | …01110100111001101100111 |
3 | 2122020002012110020212000022 |
4 | 10210232211232213031213 |
5 | 10114040244222311142 |
6 | 110441114151505355 |
7 | 4144225005241331 |
oct | 444564556471547 |
9 | 78202173225008 |
10 | 20116112307047 |
11 | 6456216a75979 |
12 | 230a77938825b |
13 | b2bc30c0bb73 |
14 | 4d78a7126451 |
15 | 24d3ed89c6d2 |
hex | 124ba5ba7367 |
20116112307047 has 2 divisors, whose sum is σ = 20116112307048. Its totient is φ = 20116112307046.
The previous prime is 20116112307013. The next prime is 20116112307073. The reversal of 20116112307047 is 74070321161102.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 20116112307047 - 234 = 20098932437863 is a prime.
It is a junction number, because it is equal to n+sod(n) for n = 20116112306998 and 20116112307016.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (20116112307007) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 10058056153523 + 10058056153524.
It is an arithmetic number, because the mean of its divisors is an integer number (10058056153524).
Almost surely, 220116112307047 is an apocalyptic number.
20116112307047 is a deficient number, since it is larger than the sum of its proper divisors (1).
20116112307047 is an equidigital number, since it uses as much as digits as its factorization.
20116112307047 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 14112, while the sum is 35.
Adding to 20116112307047 its reverse (74070321161102), we get a palindrome (94186433468149).
The spelling of 20116112307047 in words is "twenty trillion, one hundred sixteen billion, one hundred twelve million, three hundred seven thousand, forty-seven".
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