Base | Representation |
---|---|
bin | 10100011010000001011… |
… | …000011001100000010001 |
3 | 11222001122210021211010101 |
4 | 110122001120121200101 |
5 | 140433432434322111 |
6 | 2552115502332401 |
7 | 203213021200135 |
oct | 24320130314021 |
9 | 4861583254111 |
10 | 1402329995281 |
11 | 4a07a77a6226 |
12 | 1a794558a101 |
13 | a2315551ac7 |
14 | 4bc31c453c5 |
15 | 267279a8bc1 |
hex | 14681619811 |
1402329995281 has 2 divisors, whose sum is σ = 1402329995282. Its totient is φ = 1402329995280.
The previous prime is 1402329995267. The next prime is 1402329995327. The reversal of 1402329995281 is 1825999232041.
It is a weak prime.
It can be written as a sum of positive squares in only one way, i.e., 1298414670400 + 103915324881 = 1139480^2 + 322359^2 .
It is a cyclic number.
It is not a de Polignac number, because 1402329995281 - 25 = 1402329995249 is a prime.
It is a super-2 number, since 2×14023299952812 (a number of 25 digits) contains 22 as substring.
It is not a weakly prime, because it can be changed into another prime (1402329995201) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 701164997640 + 701164997641.
It is an arithmetic number, because the mean of its divisors is an integer number (701164997641).
Almost surely, 21402329995281 is an apocalyptic number.
It is an amenable number.
1402329995281 is a deficient number, since it is larger than the sum of its proper divisors (1).
1402329995281 is an equidigital number, since it uses as much as digits as its factorization.
1402329995281 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 2799360, while the sum is 55.
The spelling of 1402329995281 in words is "one trillion, four hundred two billion, three hundred twenty-nine million, nine hundred ninety-five thousand, two hundred eighty-one".
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