Base | Representation |
---|---|
bin | 11110110001101011011110… |
… | …000001010100111101000111 |
3 | 122202020211202021101211211021 |
4 | 132301223132001110331013 |
5 | 120220130201002143241 |
6 | 1155513213314252011 |
7 | 40340046054622423 |
oct | 3661533601247507 |
9 | 582224667354737 |
10 | 135355324256071 |
11 | 3a145963922467 |
12 | 13220928287007 |
13 | 5a6ac4c2b7b3a |
14 | 255d338563b83 |
15 | 109ad814760d1 |
hex | 7b1ade054f47 |
135355324256071 has 2 divisors, whose sum is σ = 135355324256072. Its totient is φ = 135355324256070.
The previous prime is 135355324256057. The next prime is 135355324256081. The reversal of 135355324256071 is 170652423553531.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 135355324256071 - 213 = 135355324247879 is a prime.
It is a super-3 number, since 3×1353553242560713 (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 (135355324256081) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 67677662128035 + 67677662128036.
It is an arithmetic number, because the mean of its divisors is an integer number (67677662128036).
Almost surely, 2135355324256071 is an apocalyptic number.
135355324256071 is a deficient number, since it is larger than the sum of its proper divisors (1).
135355324256071 is an equidigital number, since it uses as much as digits as its factorization.
135355324256071 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 11340000, while the sum is 52.
The spelling of 135355324256071 in words is "one hundred thirty-five trillion, three hundred fifty-five billion, three hundred twenty-four million, two hundred fifty-six thousand, seventy-one".
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