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255241025131 is a prime number
BaseRepresentation
bin1110110110110110001…
…1001111011001101011
3220101211012000112000011
43231231203033121223
513140213200301011
6313131152552351
724304052626261
oct3555543173153
9811735015004
10255241025131
11992799a4a84
12415738526b7
131b0b8b3a0b1
14c4d48d4431
15698ced0621
hex3b6d8cf66b

255241025131 has 2 divisors, whose sum is σ = 255241025132. Its totient is φ = 255241025130.

The previous prime is 255241025101. The next prime is 255241025153. The reversal of 255241025131 is 131520142552.

It is a strong prime.

It is a cyclic number.

It is not a de Polignac number, because 255241025131 - 27 = 255241025003 is a prime.

It is a super-2 number, since 2×2552410251312 (a number of 24 digits) contains 22 as substring.

It is a junction number, because it is equal to n+sod(n) for n = 255241025093 and 255241025102.

It is not a weakly prime, because it can be changed into another prime (255241025101) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 127620512565 + 127620512566.

It is an arithmetic number, because the mean of its divisors is an integer number (127620512566).

Almost surely, 2255241025131 is an apocalyptic number.

255241025131 is a deficient number, since it is larger than the sum of its proper divisors (1).

255241025131 is an equidigital number, since it uses as much as digits as its factorization.

255241025131 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 12000, while the sum is 31.

Adding to 255241025131 its reverse (131520142552), we get a palindrome (386761167683).

The spelling of 255241025131 in words is "two hundred fifty-five billion, two hundred forty-one million, twenty-five thousand, one hundred thirty-one".