14112520232537 has 2 divisors, whose sum is σ = 14112520232538. Its totient is φ = 14112520232536.

The previous prime is 14112520232483. The next prime is 14112520232539. The reversal of 14112520232537 is 73523202521141.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 13098398727241 + 1014121505296 = 3619171^2 + 1007036^2 .

It is a cyclic number.

It is not a de Polignac number, because 14112520232537 - 2⁸ = 14112520232281 is a prime.

Together with 14112520232539, 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 14112520232537.

It is not a weakly prime, because it can be changed into another prime (14112520232539) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 7056260116268 + 7056260116269.

It is an arithmetic number, because the mean of its divisors is an integer number (7056260116269).

Almost surely, 2^{14112520232537} is an apocalyptic number.

It is an amenable number.

14112520232537 is a deficient number, since it is larger than the sum of its proper divisors (1).

14112520232537 is an equidigital number, since it uses as much as digits as its factorization.

14112520232537 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 100800, while the sum is 38.

Adding to 14112520232537 its reverse (73523202521141), we get a palindrome (87635722753678).

The spelling of 14112520232537 in words is "fourteen trillion, one hundred twelve billion, five hundred twenty million, two hundred thirty-two thousand, five hundred thirty-seven".