Base | Representation |
---|---|
bin | 111010001111100100011011… |
… | …0100010011111000110001111 |
3 | 2111011221122221001021001110121 |
4 | 1310133020312202133012033 |
5 | 1014122214420211331112 |
6 | 5013333234204445411 |
7 | 212624256533232661 |
oct | 16437106642370617 |
9 | 2434848831231417 |
10 | 512313204011407 |
11 | 13926989a761097 |
12 | 49561893549867 |
13 | 18cb2c0a613c76 |
14 | 907213ca50531 |
15 | 3e366972cc407 |
hex | 1d1f23689f18f |
512313204011407 has 2 divisors, whose sum is σ = 512313204011408. Its totient is φ = 512313204011406.
The previous prime is 512313204011363. The next prime is 512313204011443. The reversal of 512313204011407 is 704110402313215.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 512313204011407 - 223 = 512313195622799 is a prime.
It is a super-2 number, since 2×5123132040114072 (a number of 30 digits) contains 22 as substring.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (512313204011467) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 256156602005703 + 256156602005704.
It is an arithmetic number, because the mean of its divisors is an integer number (256156602005704).
Almost surely, 2512313204011407 is an apocalyptic number.
512313204011407 is a deficient number, since it is larger than the sum of its proper divisors (1).
512313204011407 is an equidigital number, since it uses as much as digits as its factorization.
512313204011407 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 20160, while the sum is 34.
The spelling of 512313204011407 in words is "five hundred twelve trillion, three hundred thirteen billion, two hundred four million, eleven thousand, four hundred seven".
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