Base | Representation |
---|---|
bin | 11101000110110110011… |
… | …01010110000010111011 |
3 | 10112121110101122011022001 |
4 | 32203123031112002323 |
5 | 112341211142334042 |
6 | 2043235542441431 |
7 | 132153442523644 |
oct | 16433315260273 |
9 | 3477411564261 |
10 | 1000110121147 |
11 | 3561650aa605 |
12 | 1419b2a33277 |
13 | 73405044097 |
14 | 36596d106cb |
15 | 1b036269bb7 |
hex | e8db3560bb |
1000110121147 has 2 divisors, whose sum is σ = 1000110121148. Its totient is φ = 1000110121146.
The previous prime is 1000110121133. The next prime is 1000110121163. The reversal of 1000110121147 is 7411210110001.
It is a weak prime.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-1000110121147 is a prime.
It is not a weakly prime, because it can be changed into another prime (1000110121117) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 500055060573 + 500055060574.
It is an arithmetic number, because the mean of its divisors is an integer number (500055060574).
Almost surely, 21000110121147 is an apocalyptic number.
1000110121147 is a deficient number, since it is larger than the sum of its proper divisors (1).
1000110121147 is an equidigital number, since it uses as much as digits as its factorization.
1000110121147 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 56, while the sum is 19.
Adding to 1000110121147 its reverse (7411210110001), we get a palindrome (8411320231148).
The spelling of 1000110121147 in words is "one trillion, one hundred ten million, one hundred twenty-one thousand, one hundred forty-seven".
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