Base | Representation |
---|---|
bin | 10111011011000000100100… |
… | …001101011100000110010011 |
3 | 111111201202122220002020201222 |
4 | 113123000210031130012103 |
5 | 102000213214400130003 |
6 | 1003030351504515255 |
7 | 30461210110351625 |
oct | 2733004415340623 |
9 | 444652586066658 |
10 | 103011103130003 |
11 | 2a90584424a852 |
12 | b6782aa01bb2b |
13 | 4562ba006b853 |
14 | 1b61a9a582c15 |
15 | bd984db74c38 |
hex | 5db02435c193 |
103011103130003 has 2 divisors, whose sum is σ = 103011103130004. Its totient is φ = 103011103130002.
The previous prime is 103011103129999. The next prime is 103011103130011. The reversal of 103011103130003 is 300031301110301.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 103011103130003 - 22 = 103011103129999 is a prime.
It is a super-2 number, since 2×1030111031300032 (a number of 29 digits) contains 22 as substring.
It is a Sophie Germain prime.
It is not a weakly prime, because it can be changed into another prime (103011103190003) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 51505551565001 + 51505551565002.
It is an arithmetic number, because the mean of its divisors is an integer number (51505551565002).
Almost surely, 2103011103130003 is an apocalyptic number.
103011103130003 is a deficient number, since it is larger than the sum of its proper divisors (1).
103011103130003 is an equidigital number, since it uses as much as digits as its factorization.
103011103130003 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 81, while the sum is 17.
Adding to 103011103130003 its reverse (300031301110301), we get a palindrome (403042404240304).
The spelling of 103011103130003 in words is "one hundred three trillion, eleven billion, one hundred three million, one hundred thirty thousand, three".
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