Base | Representation |
---|---|
bin | 10011000111001100110… |
… | …111000100000110011111 |
3 | 11122120010020121012120022 |
4 | 103013030313010012133 |
5 | 133004321403341223 |
6 | 2443211335142355 |
7 | 163614216621011 |
oct | 23071467040637 |
9 | 4576106535508 |
10 | 1313402012063 |
11 | 467013217219 |
12 | 1926678149bb |
13 | 96b128634ab |
14 | 477d74801b1 |
15 | 242707d6ac8 |
hex | 131ccdc419f |
1313402012063 has 2 divisors, whose sum is σ = 1313402012064. Its totient is φ = 1313402012062.
The previous prime is 1313402012003. The next prime is 1313402012093. The reversal of 1313402012063 is 3602102043131.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 1313402012063 - 214 = 1313401995679 is a prime.
It is a super-2 number, since 2×13134020120632 (a number of 25 digits) contains 22 as substring.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (1313402012003) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 656701006031 + 656701006032.
It is an arithmetic number, because the mean of its divisors is an integer number (656701006032).
Almost surely, 21313402012063 is an apocalyptic number.
1313402012063 is a deficient number, since it is larger than the sum of its proper divisors (1).
1313402012063 is an equidigital number, since it uses as much as digits as its factorization.
1313402012063 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 2592, while the sum is 26.
Adding to 1313402012063 its reverse (3602102043131), we get a palindrome (4915504055194).
The spelling of 1313402012063 in words is "one trillion, three hundred thirteen billion, four hundred two million, twelve thousand, sixty-three".
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