Base | Representation |
---|---|
bin | 10011101110001011011… |
… | …000010110001100111011 |
3 | 11210120010220211110002011 |
4 | 103232023120112030323 |
5 | 134201024244114114 |
6 | 2514332241350351 |
7 | 166625302252324 |
oct | 23561330261473 |
9 | 4716126743064 |
10 | 1355253113659 |
11 | 48283a062295 |
12 | 19a7a76383b7 |
13 | 9aa52288733 |
14 | 4984782744b |
15 | 253bea6d1c4 |
hex | 13b8b61633b |
1355253113659 has 2 divisors, whose sum is σ = 1355253113660. Its totient is φ = 1355253113658.
The previous prime is 1355253113651. The next prime is 1355253113701. The reversal of 1355253113659 is 9563113525531.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 1355253113659 - 23 = 1355253113651 is a prime.
It is a super-3 number, since 3×13552531136593 (a number of 37 digits) contains 333 as substring. Note that it is a super-d number also for d = 2.
It is not a weakly prime, because it can be changed into another prime (1355253113651) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 677626556829 + 677626556830.
It is an arithmetic number, because the mean of its divisors is an integer number (677626556830).
Almost surely, 21355253113659 is an apocalyptic number.
1355253113659 is a deficient number, since it is larger than the sum of its proper divisors (1).
1355253113659 is an equidigital number, since it uses as much as digits as its factorization.
1355253113659 is an evil number, because the sum of its binary digits is even.
The product of its digits is 1822500, while the sum is 49.
The spelling of 1355253113659 in words is "one trillion, three hundred fifty-five billion, two hundred fifty-three million, one hundred thirteen thousand, six hundred fifty-nine".
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