Base | Representation |
---|---|
bin | 10000110000000011001… |
… | …010111011011110000011 |
3 | 11002001012110001220020121 |
4 | 100300003022323132003 |
5 | 122324430213311003 |
6 | 2240450550521111 |
7 | 146110304142451 |
oct | 20600312733603 |
9 | 4061173056217 |
10 | 1151104432003 |
11 | 4041a9496169 |
12 | 167112666197 |
13 | 847195bb056 |
14 | 3d9dc76b9d1 |
15 | 1ee222a4cbd |
hex | 10c032bb783 |
1151104432003 has 2 divisors, whose sum is σ = 1151104432004. Its totient is φ = 1151104432002.
The previous prime is 1151104431989. The next prime is 1151104432013. The reversal of 1151104432003 is 3002344011511.
It is a strong prime.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-1151104432003 is a prime.
It is not a weakly prime, because it can be changed into another prime (1151104432013) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 575552216001 + 575552216002.
It is an arithmetic number, because the mean of its divisors is an integer number (575552216002).
Almost surely, 21151104432003 is an apocalyptic number.
1151104432003 is a deficient number, since it is larger than the sum of its proper divisors (1).
1151104432003 is an equidigital number, since it uses as much as digits as its factorization.
1151104432003 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 1440, while the sum is 25.
Adding to 1151104432003 its reverse (3002344011511), we get a palindrome (4153448443514).
The spelling of 1151104432003 in words is "one trillion, one hundred fifty-one billion, one hundred four million, four hundred thirty-two thousand, three".
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