Base | Representation |
---|---|
bin | 10000011101100010001… |
… | …001010011000010010011 |
3 | 11000010212212122200012202 |
4 | 100131202021103002103 |
5 | 122013221042101113 |
6 | 2223402054513415 |
7 | 144504521606432 |
oct | 20354211230223 |
9 | 4003785580182 |
10 | 1131223003283 |
11 | 3a682699253a |
12 | 1632a436586b |
13 | 8289b659c49 |
14 | 3ca7410d119 |
15 | 1e65bb5b258 |
hex | 10762253093 |
1131223003283 has 2 divisors, whose sum is σ = 1131223003284. Its totient is φ = 1131223003282.
The previous prime is 1131223003103. The next prime is 1131223003321. The reversal of 1131223003283 is 3823003221311.
It is a strong prime.
It is an emirp because it is prime and its reverse (3823003221311) is a distict prime.
It is a cyclic number.
It is not a de Polignac number, because 1131223003283 - 214 = 1131222986899 is a prime.
It is a super-2 number, since 2×11312230032832 (a number of 25 digits) contains 22 as substring.
It is not a weakly prime, because it can be changed into another prime (1131225003283) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 565611501641 + 565611501642.
It is an arithmetic number, because the mean of its divisors is an integer number (565611501642).
Almost surely, 21131223003283 is an apocalyptic number.
1131223003283 is a deficient number, since it is larger than the sum of its proper divisors (1).
1131223003283 is an equidigital number, since it uses as much as digits as its factorization.
1131223003283 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 5184, while the sum is 29.
Adding to 1131223003283 its reverse (3823003221311), we get a palindrome (4954226224594).
The spelling of 1131223003283 in words is "one trillion, one hundred thirty-one billion, two hundred twenty-three million, three thousand, two hundred eighty-three".
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