Base | Representation |
---|---|
bin | 101000010010000111110111… |
… | …0110001111000001101011001 |
3 | 1201110121001022110000220121112 |
4 | 1100210033232301320031121 |
5 | 332420404040312133101 |
6 | 3253335010022150105 |
7 | 134430552356643665 |
oct | 12044175661701531 |
9 | 1643531273026545 |
10 | 354334513005401 |
11 | a29a1472123235 |
12 | 338a849b582335 |
13 | 122937a7ac54b6 |
14 | 636dc4c05d8a5 |
15 | 2ae70c1a6d6bb |
hex | 14243eec78359 |
354334513005401 has 2 divisors, whose sum is σ = 354334513005402. Its totient is φ = 354334513005400.
The previous prime is 354334513005391. The next prime is 354334513005467. The reversal of 354334513005401 is 104500315433453.
It is a weak prime.
It can be written as a sum of positive squares in only one way, i.e., 255886444595401 + 98448068410000 = 15996451^2 + 9922100^2 .
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-354334513005401 is a prime.
It is not a weakly prime, because it can be changed into another prime (354334513002401) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 177167256502700 + 177167256502701.
It is an arithmetic number, because the mean of its divisors is an integer number (177167256502701).
Almost surely, 2354334513005401 is an apocalyptic number.
It is an amenable number.
354334513005401 is a deficient number, since it is larger than the sum of its proper divisors (1).
354334513005401 is an equidigital number, since it uses as much as digits as its factorization.
354334513005401 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 648000, while the sum is 41.
Adding to 354334513005401 its reverse (104500315433453), we get a palindrome (458834828438854).
The spelling of 354334513005401 in words is "three hundred fifty-four trillion, three hundred thirty-four billion, five hundred thirteen million, five thousand, four hundred one".
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