10112513141 has 2 divisors, whose sum is σ = 10112513142. Its totient is φ = 10112513140.

The previous prime is 10112513137. The next prime is 10112513143. The reversal of 10112513141 is 14131521101.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 5726100241 + 4386412900 = 75671^2 + 66230^2 .

It is a cyclic number.

It is not a de Polignac number, because 10112513141 - 2² = 10112513137 is a prime.

Together with 10112513143, it forms a pair of twin primes.

It is a Chen prime.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (10112513143) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 5056256570 + 5056256571.

It is an arithmetic number, because the mean of its divisors is an integer number (5056256571).

Almost surely, 2^10112513141 is an apocalyptic number.

It is an amenable number.

10112513141 is a deficient number, since it is larger than the sum of its proper divisors (1).

10112513141 is an equidigital number, since it uses as much as digits as its factorization.

10112513141 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 120, while the sum is 20.

It can be divided in two parts, 1011 and 2513141, that added together give a palindrome (2514152).

The spelling of 10112513141 in words is "ten billion, one hundred twelve million, five hundred thirteen thousand, one hundred forty-one".