Base | Representation |
---|---|
bin | 10010001111010000111… |
… | …001001110011101001011 |
3 | 11102211002102121020211021 |
4 | 102033100321032131023 |
5 | 131013320034123413 |
6 | 2355435421013311 |
7 | 156356636313166 |
oct | 22172071163513 |
9 | 4384072536737 |
10 | 1253340145483 |
11 | 4435a1923958 |
12 | 182aa50b4237 |
13 | 9126035487a |
14 | 4493a6c4add |
15 | 2290791318d |
hex | 123d0e4e74b |
1253340145483 has 2 divisors, whose sum is σ = 1253340145484. Its totient is φ = 1253340145482.
The previous prime is 1253340145451. The next prime is 1253340145559. The reversal of 1253340145483 is 3845410433521.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 1253340145483 - 25 = 1253340145451 is a prime.
It is not a weakly prime, because it can be changed into another prime (1253340145433) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 626670072741 + 626670072742.
It is an arithmetic number, because the mean of its divisors is an integer number (626670072742).
Almost surely, 21253340145483 is an apocalyptic number.
1253340145483 is a deficient number, since it is larger than the sum of its proper divisors (1).
1253340145483 is an equidigital number, since it uses as much as digits as its factorization.
1253340145483 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 691200, while the sum is 43.
The spelling of 1253340145483 in words is "one trillion, two hundred fifty-three billion, three hundred forty million, one hundred forty-five thousand, four hundred eighty-three".
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