Base | Representation |
---|---|
bin | 11011100010110110011001… |
… | …110111101010100110100111 |
3 | 120212221002000121120100111122 |
4 | 123202312121313222212213 |
5 | 111334243042311341101 |
6 | 1105351532250124155 |
7 | 34342142004601526 |
oct | 3342663167524647 |
9 | 525832017510448 |
10 | 121142134090151 |
11 | 356660aa585809 |
12 | 117061a772065b |
13 | 527986744cc25 |
14 | 21cb44c3064bd |
15 | e012b913b51b |
hex | 6e2d99dea9a7 |
121142134090151 has 2 divisors, whose sum is σ = 121142134090152. Its totient is φ = 121142134090150.
The previous prime is 121142134090127. The next prime is 121142134090171. The reversal of 121142134090151 is 151090431241121.
It is a strong prime.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-121142134090151 is a prime.
It is a super-3 number, since 3×1211421340901513 (a number of 43 digits) contains 333 as substring. Note that it is a super-d number also for d = 2.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (121142134090111) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 60571067045075 + 60571067045076.
It is an arithmetic number, because the mean of its divisors is an integer number (60571067045076).
Almost surely, 2121142134090151 is an apocalyptic number.
121142134090151 is a deficient number, since it is larger than the sum of its proper divisors (1).
121142134090151 is an equidigital number, since it uses as much as digits as its factorization.
121142134090151 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 8640, while the sum is 35.
The spelling of 121142134090151 in words is "one hundred twenty-one trillion, one hundred forty-two billion, one hundred thirty-four million, ninety thousand, one hundred fifty-one".
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