Base | Representation |
---|---|
bin | 11000100111010001… |
… | …01001010110100011 |
3 | 1021002221002012220021 |
4 | 30103220221112203 |
5 | 204030331333011 |
6 | 10023124454311 |
7 | 645314621221 |
oct | 142350512643 |
9 | 37087065807 |
10 | 13214324131 |
11 | 5671161338 |
12 | 268954a397 |
13 | 1327902b08 |
14 | 8d4dd9b11 |
15 | 52518a471 |
hex | 313a295a3 |
13214324131 has 2 divisors, whose sum is σ = 13214324132. Its totient is φ = 13214324130.
The previous prime is 13214324129. The next prime is 13214324141. The reversal of 13214324131 is 13142341231.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 13214324131 - 21 = 13214324129 is a prime.
It is a super-2 number, since 2×132143241312 (a number of 21 digits) contains 22 as substring.
Together with 13214324129, it forms a pair of twin primes.
It is a junction number, because it is equal to n+sod(n) for n = 13214324096 and 13214324105.
It is not a weakly prime, because it can be changed into another prime (13214324141) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6607162065 + 6607162066.
It is an arithmetic number, because the mean of its divisors is an integer number (6607162066).
Almost surely, 213214324131 is an apocalyptic number.
13214324131 is a deficient number, since it is larger than the sum of its proper divisors (1).
13214324131 is an equidigital number, since it uses as much as digits as its factorization.
13214324131 is an evil number, because the sum of its binary digits is even.
The product of its digits is 1728, while the sum is 25.
Adding to 13214324131 its reverse (13142341231), we get a palindrome (26356665362).
The spelling of 13214324131 in words is "thirteen billion, two hundred fourteen million, three hundred twenty-four thousand, one hundred thirty-one".
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