483560131 has 2 divisors, whose sum is σ = 483560132. Its totient is φ = 483560130.

The previous prime is 483560101. The next prime is 483560159. The reversal of 483560131 is 131065384.

It is a strong prime.

It is a cyclic number.

It is not a de Polignac number, because 483560131 - 2²³ = 475171523 is a prime.

It is a super-2 number, since 2×483560131² = 467660800585474322, which contains 22 as substring.

It is a junction number, because it is equal to n+sod(n) for n = 483560093 and 483560102.

It is not a weakly prime, because it can be changed into another prime (483560101) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 241780065 + 241780066.

It is an arithmetic number, because the mean of its divisors is an integer number (241780066).

Almost surely, 2^483560131 is an apocalyptic number.

483560131 is a deficient number, since it is larger than the sum of its proper divisors (1).

483560131 is an equidigital number, since it uses as much as digits as its factorization.

483560131 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 8640, while the sum is 31.

The square root of 483560131 is about 21990.0007048658. Note that the first 3 decimals coincide. The cubic root of 483560131 is about 784.9045178469.

Subtracting from 483560131 its sum of digits (31), we obtain a square (483560100 = 21990²).

The spelling of 483560131 in words is "four hundred eighty-three million, five hundred sixty thousand, one hundred thirty-one".