Base | Representation |
---|---|
bin | 1011111011001000000011… |
… | …0100011100110000100011 |
3 | 1201102100020102020202210221 |
4 | 2332302000310130300203 |
5 | 3204300100230422243 |
6 | 43514500230103511 |
7 | 2522123653240444 |
oct | 276620064346043 |
9 | 51370212222727 |
10 | 13110401420323 |
11 | 41a50a98a5939 |
12 | 1578a72aaab97 |
13 | 7413c83a322c |
14 | 33479265d2cb |
15 | 17b071b23bed |
hex | bec80d1cc23 |
13110401420323 has 2 divisors, whose sum is σ = 13110401420324. Its totient is φ = 13110401420322.
The previous prime is 13110401420297. The next prime is 13110401420387. The reversal of 13110401420323 is 32302410401131.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 13110401420323 - 213 = 13110401412131 is a prime.
It is a super-3 number, since 3×131104014203233 (a number of 40 digits) contains 333 as substring.
It is not a weakly prime, because it can be changed into another prime (13110401420723) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6555200710161 + 6555200710162.
It is an arithmetic number, because the mean of its divisors is an integer number (6555200710162).
Almost surely, 213110401420323 is an apocalyptic number.
13110401420323 is a deficient number, since it is larger than the sum of its proper divisors (1).
13110401420323 is an equidigital number, since it uses as much as digits as its factorization.
13110401420323 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 13110401420323 its reverse (32302410401131), we get a palindrome (45412811821454).
The spelling of 13110401420323 in words is "thirteen trillion, one hundred ten billion, four hundred one million, four hundred twenty thousand, three hundred twenty-three".
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