Base | Representation |
---|---|
bin | 100111011001001001000… |
… | …1111100101011100000011 |
3 | 201011120202210120020221211 |
4 | 1032302102033211130003 |
5 | 1202201100230231003 |
6 | 15303113041230551 |
7 | 1066104515450425 |
oct | 116622217453403 |
9 | 21146683506854 |
10 | 5414112352003 |
11 | 17a8126305893 |
12 | 73535b332457 |
13 | 30371938b244 |
14 | 14a089d0a815 |
15 | 95c7804296d |
hex | 4ec923e5703 |
5414112352003 has 2 divisors, whose sum is σ = 5414112352004. Its totient is φ = 5414112352002.
The previous prime is 5414112351961. The next prime is 5414112352067. The reversal of 5414112352003 is 3002532114145.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 5414112352003 - 217 = 5414112220931 is a prime.
It is a super-3 number, since 3×54141123520033 (a number of 39 digits) contains 333 as substring. Note that it is a super-d number also for d = 2.
It is not a weakly prime, because it can be changed into another prime (5414112352603) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 2707056176001 + 2707056176002.
It is an arithmetic number, because the mean of its divisors is an integer number (2707056176002).
Almost surely, 25414112352003 is an apocalyptic number.
5414112352003 is a deficient number, since it is larger than the sum of its proper divisors (1).
5414112352003 is an equidigital number, since it uses as much as digits as its factorization.
5414112352003 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 14400, while the sum is 31.
Adding to 5414112352003 its reverse (3002532114145), we get a palindrome (8416644466148).
The spelling of 5414112352003 in words is "five trillion, four hundred fourteen billion, one hundred twelve million, three hundred fifty-two thousand, three".
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