Base | Representation |
---|---|
bin | 1010011101101011011101… |
… | …1100100010011000000011 |
3 | 1111201212102020111022111211 |
4 | 2213122313130202120003 |
5 | 3001444220013130201 |
6 | 40245154000550551 |
7 | 2265131425361431 |
oct | 247326734423003 |
9 | 44655366438454 |
10 | 11505000130051 |
11 | 373627470309a |
12 | 13598b5322457 |
13 | 655bbc44795c |
14 | 2babb8891d51 |
15 | 14e41126c651 |
hex | a76b7722603 |
11505000130051 has 2 divisors, whose sum is σ = 11505000130052. Its totient is φ = 11505000130050.
The previous prime is 11505000130033. The next prime is 11505000130073. The reversal of 11505000130051 is 15003100050511.
11505000130051 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.
It is an a-pointer prime, because the next prime (11505000130073) can be obtained adding 11505000130051 to its sum of digits (22).
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 11505000130051 - 219 = 11504999605763 is a prime.
It is not a weakly prime, because it can be changed into another prime (11505000130031) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 5752500065025 + 5752500065026.
It is an arithmetic number, because the mean of its divisors is an integer number (5752500065026).
Almost surely, 211505000130051 is an apocalyptic number.
11505000130051 is a deficient number, since it is larger than the sum of its proper divisors (1).
11505000130051 is an equidigital number, since it uses as much as digits as its factorization.
11505000130051 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 375, while the sum is 22.
Adding to 11505000130051 its reverse (15003100050511), we get a palindrome (26508100180562).
The spelling of 11505000130051 in words is "eleven trillion, five hundred five billion, one hundred thirty thousand, fifty-one".
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