Base | Representation |
---|---|
bin | 1100001110100000001111… |
… | …0011001111010100110001 |
3 | 1202121011112111002222221011 |
4 | 3003220003303033110301 |
5 | 3230223400204301431 |
6 | 44331434430230521 |
7 | 2555150534336332 |
oct | 303500363172461 |
9 | 52534474088834 |
10 | 13443311400241 |
11 | 4313304901148 |
12 | 16114a1955441 |
13 | 766911420328 |
14 | 34693449c289 |
15 | 184a5861ccb1 |
hex | c3a03ccf531 |
13443311400241 has 2 divisors, whose sum is σ = 13443311400242. Its totient is φ = 13443311400240.
The previous prime is 13443311400221. The next prime is 13443311400271. The reversal of 13443311400241 is 14200411334431.
It is a weak prime.
It can be written as a sum of positive squares in only one way, i.e., 12221988032016 + 1221323368225 = 3495996^2 + 1105135^2 .
It is a cyclic number.
It is not a de Polignac number, because 13443311400241 - 25 = 13443311400209 is a prime.
It is not a weakly prime, because it can be changed into another prime (13443311400221) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6721655700120 + 6721655700121.
It is an arithmetic number, because the mean of its divisors is an integer number (6721655700121).
Almost surely, 213443311400241 is an apocalyptic number.
It is an amenable number.
13443311400241 is a deficient number, since it is larger than the sum of its proper divisors (1).
13443311400241 is an equidigital number, since it uses as much as digits as its factorization.
13443311400241 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 13824, while the sum is 31.
Adding to 13443311400241 its reverse (14200411334431), we get a palindrome (27643722734672).
The spelling of 13443311400241 in words is "thirteen trillion, four hundred forty-three billion, three hundred eleven million, four hundred thousand, two hundred forty-one".
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