101152433117 has 2 divisors, whose sum is σ = 101152433118. Its totient is φ = 101152433116.

The previous prime is 101152433099. The next prime is 101152433119. The reversal of 101152433117 is 711334251101.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 99236970361 + 1915462756 = 315019^2 + 43766^2 .

It is a cyclic number.

It is not a de Polignac number, because 101152433117 - 2⁸ = 101152432861 is a prime.

Together with 101152433119, it forms a pair of twin primes.

It is a Chen prime.

It is a self number, because there is not a number n which added to its sum of digits gives 101152433117.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (101152433119) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 50576216558 + 50576216559.

It is an arithmetic number, because the mean of its divisors is an integer number (50576216559).

Almost surely, 2^101152433117 is an apocalyptic number.

It is an amenable number.

101152433117 is a deficient number, since it is larger than the sum of its proper divisors (1).

101152433117 is an equidigital number, since it uses as much as digits as its factorization.

101152433117 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 2520, while the sum is 29.

Adding to 101152433117 its reverse (711334251101), we get a palindrome (812486684218).

The spelling of 101152433117 in words is "one hundred one billion, one hundred fifty-two million, four hundred thirty-three thousand, one hundred seventeen".