Base | Representation |
---|---|
bin | 1010000000100100100011… |
… | …1100111000100000110001 |
3 | 1102222001122111120100000112 |
4 | 2200021020330320200301 |
5 | 2420301044103211032 |
6 | 35223332313231105 |
7 | 2214036266464133 |
oct | 240111074704061 |
9 | 42861574510015 |
10 | 11004930132017 |
11 | 356318a186088 |
12 | 12989b4a02495 |
13 | 61a9b695c233 |
14 | 2a08da24b053 |
15 | 1413e440e1b2 |
hex | a0248f38831 |
11004930132017 has 2 divisors, whose sum is σ = 11004930132018. Its totient is φ = 11004930132016.
The previous prime is 11004930132011. The next prime is 11004930132043. The reversal of 11004930132017 is 71023103940011.
It is a weak prime.
It can be written as a sum of positive squares in only one way, i.e., 11004671694241 + 258437776 = 3317329^2 + 16076^2 .
It is a cyclic number.
It is not a de Polignac number, because 11004930132017 - 210 = 11004930130993 is a prime.
It is a super-2 number, since 2×110049301320172 (a number of 27 digits) contains 22 as substring.
It is not a weakly prime, because it can be changed into another prime (11004930132011) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 5502465066008 + 5502465066009.
It is an arithmetic number, because the mean of its divisors is an integer number (5502465066009).
Almost surely, 211004930132017 is an apocalyptic number.
It is an amenable number.
11004930132017 is a deficient number, since it is larger than the sum of its proper divisors (1).
11004930132017 is an equidigital number, since it uses as much as digits as its factorization.
11004930132017 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 4536, while the sum is 32.
The spelling of 11004930132017 in words is "eleven trillion, four billion, nine hundred thirty million, one hundred thirty-two thousand, seventeen".
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