Base | Representation |
---|---|
bin | 110001011111101000001… |
… | …010100001010011101001 |
3 | 110001011010110120112222121 |
4 | 301133220022201103221 |
5 | 421211141313444423 |
6 | 11122255104151241 |
7 | 500505154466626 |
oct | 61375012412351 |
9 | 13034113515877 |
10 | 3401214203113 |
11 | 10a14a0825921 |
12 | 46b217908521 |
13 | 1b896b89640c |
14 | ba89644534d |
15 | 5d717a3145d |
hex | 317e82a14e9 |
3401214203113 has 2 divisors, whose sum is σ = 3401214203114. Its totient is φ = 3401214203112.
The previous prime is 3401214203059. The next prime is 3401214203137. The reversal of 3401214203113 is 3113024121043.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 2122840258009 + 1278373945104 = 1456997^2 + 1130652^2 .
It is an emirp because it is prime and its reverse (3113024121043) is a distict prime.
It is a cyclic number.
It is not a de Polignac number, because 3401214203113 - 29 = 3401214202601 is a prime.
It is not a weakly prime, because it can be changed into another prime (3401214203143) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 1700607101556 + 1700607101557.
It is an arithmetic number, because the mean of its divisors is an integer number (1700607101557).
Almost surely, 23401214203113 is an apocalyptic number.
It is an amenable number.
3401214203113 is a deficient number, since it is larger than the sum of its proper divisors (1).
3401214203113 is an equidigital number, since it uses as much as digits as its factorization.
3401214203113 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 1728, while the sum is 25.
Adding to 3401214203113 its reverse (3113024121043), we get a palindrome (6514238324156).
The spelling of 3401214203113 in words is "three trillion, four hundred one billion, two hundred fourteen million, two hundred three thousand, one hundred thirteen".
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