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505231012434131 is a prime number
BaseRepresentation
bin111001011100000010100001…
…1000001101001000011010011
32110020212121112221100100020122
41302320011003001221003103
51012210201103140343011
64550311531204300455
7211263510322131644
oct16270050301510323
92406777487310218
10505231012434131
11136a892a61a0285
12487b91a39a172b
13188bb106b10b59
148ca94509956cb
153d623410083db
hex1cb81430690d3

505231012434131 has 2 divisors, whose sum is σ = 505231012434132. Its totient is φ = 505231012434130.

The previous prime is 505231012433989. The next prime is 505231012434157. The reversal of 505231012434131 is 131434210132505.

It is a strong prime.

It is a cyclic number.

It is not a de Polignac number, because 505231012434131 - 242 = 500832965923027 is a prime.

It is a super-4 number, since 4×5052310124341314 (a number of 60 digits) contains 4444 as substring. Note that it is a super-d number also for d = 2.

It is a junction number, because it is equal to n+sod(n) for n = 505231012434091 and 505231012434100.

It is not a weakly prime, because it can be changed into another prime (505231012434331) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 252615506217065 + 252615506217066.

It is an arithmetic number, because the mean of its divisors is an integer number (252615506217066).

Almost surely, 2505231012434131 is an apocalyptic number.

505231012434131 is a deficient number, since it is larger than the sum of its proper divisors (1).

505231012434131 is an equidigital number, since it uses as much as digits as its factorization.

505231012434131 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 43200, while the sum is 35.

Adding to 505231012434131 its reverse (131434210132505), we get a palindrome (636665222566636).

The spelling of 505231012434131 in words is "five hundred five trillion, two hundred thirty-one billion, twelve million, four hundred thirty-four thousand, one hundred thirty-one".