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511030213014241 is a prime number
BaseRepresentation
bin111010000110001110111111…
…0010100001111001011100001
32111000102000022012100110201101
41310030131332110033023201
51013440204334012423431
65010512012223034401
7212432500033546654
oct16414357624171341
92430360265313641
10511030213014241
11138914772797538
12493950b443ba01
1318c1bc3334295b
149029dcc44989b
153e1310675b961
hex1d0c77e50f2e1

511030213014241 has 2 divisors, whose sum is σ = 511030213014242. Its totient is φ = 511030213014240.

The previous prime is 511030213014229. The next prime is 511030213014263. The reversal of 511030213014241 is 142410312030115.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 413264175210000 + 97766037804241 = 20328900^2 + 9887671^2 .

It is a cyclic number.

It is not a de Polignac number, because 511030213014241 - 225 = 511030179459809 is a prime.

It is a super-2 number, since 2×5110302130142412 (a number of 30 digits) contains 22 as substring.

It is not a weakly prime, because it can be changed into another prime (511030213017241) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 255515106507120 + 255515106507121.

It is an arithmetic number, because the mean of its divisors is an integer number (255515106507121).

Almost surely, 2511030213014241 is an apocalyptic number.

It is an amenable number.

511030213014241 is a deficient number, since it is larger than the sum of its proper divisors (1).

511030213014241 is an equidigital number, since it uses as much as digits as its factorization.

511030213014241 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 2880, while the sum is 28.

Adding to 511030213014241 its reverse (142410312030115), we get a palindrome (653440525044356).

The spelling of 511030213014241 in words is "five hundred eleven trillion, thirty billion, two hundred thirteen million, fourteen thousand, two hundred forty-one".