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510212115144131 is a prime number
BaseRepresentation
bin111010000000010010000001…
…1111000100001110111000011
32110220111210122112212200011012
41310000210003320201313003
51013333303402434103011
65005044113202311135
7212316415562441141
oct16400440370416703
92426453575780135
10510212115144131
111386298198840a4
12492826390584ab
1318b8ca46ab97ca
148ddc58069c591
153debbc46cca8b
hex1d00903e21dc3

510212115144131 has 2 divisors, whose sum is σ = 510212115144132. Its totient is φ = 510212115144130.

The previous prime is 510212115144119. The next prime is 510212115144139. The reversal of 510212115144131 is 131441511212015.

It is a strong prime.

It is a cyclic number.

It is not a de Polignac number, because 510212115144131 - 214 = 510212115127747 is a prime.

It is a super-2 number, since 2×5102121151441312 (a number of 30 digits) contains 22 as substring.

It is not a weakly prime, because it can be changed into another prime (510212115144139) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 255106057572065 + 255106057572066.

It is an arithmetic number, because the mean of its divisors is an integer number (255106057572066).

Almost surely, 2510212115144131 is an apocalyptic number.

510212115144131 is a deficient number, since it is larger than the sum of its proper divisors (1).

510212115144131 is an equidigital number, since it uses as much as digits as its factorization.

510212115144131 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 4800, while the sum is 32.

Adding to 510212115144131 its reverse (131441511212015), we get a palindrome (641653626356146).

The spelling of 510212115144131 in words is "five hundred ten trillion, two hundred twelve billion, one hundred fifteen million, one hundred forty-four thousand, one hundred thirty-one".