1011131125031 has 2 divisors, whose sum is σ = 1011131125032. Its totient is φ = 1011131125030.

The previous prime is 1011131124947. The next prime is 1011131125033. The reversal of 1011131125031 is 1305211311101.

It is a strong prime.

It is a cyclic number.

It is not a de Polignac number, because 1011131125031 - 2²² = 1011126930727 is a prime.

It is a super-2 number, since 2×1011131125031² (a number of 25 digits) contains 22 as substring.

Together with 1011131125033, it forms a pair of twin primes.

It is a Chen prime.

It is a junction number, because it is equal to n+sod(n) for n = 1011131124994 and 1011131125012.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (1011131125033) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 505565562515 + 505565562516.

It is an arithmetic number, because the mean of its divisors is an integer number (505565562516).

Almost surely, 2^{1011131125031} is an apocalyptic number.

1011131125031 is a deficient number, since it is larger than the sum of its proper divisors (1).

1011131125031 is an equidigital number, since it uses as much as digits as its factorization.

1011131125031 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 90, while the sum is 20.

Adding to 1011131125031 its reverse (1305211311101), we get a palindrome (2316342436132).

The spelling of 1011131125031 in words is "one trillion, eleven billion, one hundred thirty-one million, one hundred twenty-five thousand, thirty-one".