11512112131111 has 2 divisors, whose sum is σ = 11512112131112. Its totient is φ = 11512112131110.

The previous prime is 11512112131097. The next prime is 11512112131171. The reversal of 11512112131111 is 11113121121511.

It is a m-pointer prime, because the next prime (11512112131171) can be obtained adding 11512112131111 to its product of digits (60).

It is a weak prime.

It is a cyclic number.

It is not a de Polignac number, because 11512112131111 - 2¹³ = 11512112122919 is a prime.

It is a super-4 number, since 4×11512112131111⁴ (a number of 53 digits) contains 4444 as substring.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (11512112131171) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 5756056065555 + 5756056065556.

It is an arithmetic number, because the mean of its divisors is an integer number (5756056065556).

Almost surely, 2^{11512112131111} is an apocalyptic number.

11512112131111 is a deficient number, since it is larger than the sum of its proper divisors (1).

11512112131111 is an equidigital number, since it uses as much as digits as its factorization.

11512112131111 is an evil number, because the sum of its binary digits is even.

The product of its digits is 60, while the sum is 22.

Adding to 11512112131111 its reverse (11113121121511), we get a palindrome (22625233252622).

The spelling of 11512112131111 in words is "eleven trillion, five hundred twelve billion, one hundred twelve million, one hundred thirty-one thousand, one hundred eleven".