Base | Representation |
---|---|
bin | 10011100010000001100… |
… | …110011001000000101011 |
3 | 11202022110111211221111121 |
4 | 103202001212121000223 |
5 | 133442313223334011 |
6 | 2504333340240111 |
7 | 165654032534326 |
oct | 23420146310053 |
9 | 4668414757447 |
10 | 1342204121131 |
11 | 478253265045 |
12 | 198165540037 |
13 | 997529c2093 |
14 | 48d6a7877bd |
15 | 24da918ee71 |
hex | 1388199902b |
1342204121131 has 2 divisors, whose sum is σ = 1342204121132. Its totient is φ = 1342204121130.
The previous prime is 1342204121107. The next prime is 1342204121161. The reversal of 1342204121131 is 1311214022431.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 1342204121131 - 227 = 1342069903403 is a prime.
It is a super-2 number, since 2×13422041211312 (a number of 25 digits) contains 22 as substring.
It is a junction number, because it is equal to n+sod(n) for n = 1342204121096 and 1342204121105.
It is not a weakly prime, because it can be changed into another prime (1342204121161) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 671102060565 + 671102060566.
It is an arithmetic number, because the mean of its divisors is an integer number (671102060566).
Almost surely, 21342204121131 is an apocalyptic number.
1342204121131 is a deficient number, since it is larger than the sum of its proper divisors (1).
1342204121131 is an equidigital number, since it uses as much as digits as its factorization.
1342204121131 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 1152, while the sum is 25.
Adding to 1342204121131 its reverse (1311214022431), we get a palindrome (2653418143562).
The spelling of 1342204121131 in words is "one trillion, three hundred forty-two billion, two hundred four million, one hundred twenty-one thousand, one hundred thirty-one".
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