101433143010131 has 2 divisors, whose sum is σ = 101433143010132. Its totient is φ = 101433143010130.

The previous prime is 101433143010113. The next prime is 101433143010149. The reversal of 101433143010131 is 131010341334101.

Together with previous prime (101433143010113) it forms an Ormiston pair, because they use the same digits, order apart.

It is a balanced prime because it is at equal distance from previous prime (101433143010113) and next prime (101433143010149).

It is an emirp because it is prime and its reverse (131010341334101) is a distict prime.

It is a cyclic number.

It is not a de Polignac number, because 101433143010131 - 2³⁰ = 101432069268307 is a prime.

It is a super-2 number, since 2×101433143010131² (a number of 29 digits) contains 22 as substring.

It is a self number, because there is not a number n which added to its sum of digits gives 101433143010131.

It is not a weakly prime, because it can be changed into another prime (101433143010181) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (23) of ones.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 50716571505065 + 50716571505066.

It is an arithmetic number, because the mean of its divisors is an integer number (50716571505066).

Almost surely, 2^{101433143010131} is an apocalyptic number.

101433143010131 is a deficient number, since it is larger than the sum of its proper divisors (1).

101433143010131 is an equidigital number, since it uses as much as digits as its factorization.

101433143010131 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 1296, while the sum is 26.

Adding to 101433143010131 its reverse (131010341334101), we get a palindrome (232443484344232).

The spelling of 101433143010131 in words is "one hundred one trillion, four hundred thirty-three billion, one hundred forty-three million, ten thousand, one hundred thirty-one".