Base | Representation |
---|---|
bin | 110001011110101011111… |
… | …001100000110000001111 |
3 | 110001001112000000202221211 |
4 | 301132223321200300033 |
5 | 421202103444031224 |
6 | 11122010505035251 |
7 | 500441126401531 |
oct | 61365371406017 |
9 | 13031460022854 |
10 | 3400203111439 |
11 | 10a1022016772 |
12 | 46ab95185b27 |
13 | 1b883a274814 |
14 | ba7dc04b851 |
15 | 5d6a8dad894 |
hex | 317abe60c0f |
3400203111439 has 2 divisors, whose sum is σ = 3400203111440. Its totient is φ = 3400203111438.
The previous prime is 3400203111407. The next prime is 3400203111479. The reversal of 3400203111439 is 9341113020043.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 3400203111439 - 25 = 3400203111407 is a prime.
It is a super-3 number, since 3×34002031114393 (a number of 39 digits) contains 333 as substring. Note that it is a super-d number also for d = 2.
It is a self number, because there is not a number n which added to its sum of digits gives 3400203111439.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (3400203111479) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 1700101555719 + 1700101555720.
It is an arithmetic number, because the mean of its divisors is an integer number (1700101555720).
Almost surely, 23400203111439 is an apocalyptic number.
3400203111439 is a deficient number, since it is larger than the sum of its proper divisors (1).
3400203111439 is an equidigital number, since it uses as much as digits as its factorization.
3400203111439 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 7776, while the sum is 31.
The spelling of 3400203111439 in words is "three trillion, four hundred billion, two hundred three million, one hundred eleven thousand, four hundred thirty-nine".
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