Base | Representation |
---|---|
bin | 101101110001111100100110… |
… | …000011001000101100111111 |
3 | 222101220100010010122201022011 |
4 | 231301330212003020230333 |
5 | 202342311323004203341 |
6 | 1552120202314232051 |
7 | 60260440116554212 |
oct | 5561744603105477 |
9 | 871810103581264 |
10 | 201344410225471 |
11 | 59177759597217 |
12 | 1a6b9a8b725627 |
13 | 884690903bc36 |
14 | 37a119b397779 |
15 | 184266bc54b81 |
hex | b71f260c8b3f |
201344410225471 has 2 divisors, whose sum is σ = 201344410225472. Its totient is φ = 201344410225470.
The previous prime is 201344410225393. The next prime is 201344410225489. The reversal of 201344410225471 is 174522014443102.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 201344410225471 - 29 = 201344410224959 is a prime.
It is a super-4 number, since 4×2013444102254714 (a number of 58 digits) contains 4444 as substring. Note that it is a super-d number also for d = 3.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (201344410222471) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 100672205112735 + 100672205112736.
It is an arithmetic number, because the mean of its divisors is an integer number (100672205112736).
Almost surely, 2201344410225471 is an apocalyptic number.
201344410225471 is a deficient number, since it is larger than the sum of its proper divisors (1).
201344410225471 is an equidigital number, since it uses as much as digits as its factorization.
201344410225471 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 215040, while the sum is 40.
Adding to 201344410225471 its reverse (174522014443102), we get a palindrome (375866424668573).
The spelling of 201344410225471 in words is "two hundred one trillion, three hundred forty-four billion, four hundred ten million, two hundred twenty-five thousand, four hundred seventy-one".
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