1123413412157 has 2 divisors, whose sum is σ = 1123413412158. Its totient is φ = 1123413412156.

The previous prime is 1123413412103. The next prime is 1123413412159. The reversal of 1123413412157 is 7512143143211.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 1093971656761 + 29441755396 = 1045931^2 + 171586^2 .

It is a cyclic number.

It is not a de Polignac number, because 1123413412157 - 2¹⁴ = 1123413395773 is a prime.

It is a super-2 number, since 2×1123413412157² (a number of 25 digits) contains 22 as substring.

Together with 1123413412159, 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 (1123413412159) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 561706706078 + 561706706079.

It is an arithmetic number, because the mean of its divisors is an integer number (561706706079).

Almost surely, 2^{1123413412157} is an apocalyptic number.

It is an amenable number.

1123413412157 is a deficient number, since it is larger than the sum of its proper divisors (1).

1123413412157 is an equidigital number, since it uses as much as digits as its factorization.

1123413412157 is an evil number, because the sum of its binary digits is even.

The product of its digits is 20160, while the sum is 35.

Adding to 1123413412157 its reverse (7512143143211), we get a palindrome (8635556555368).

The spelling of 1123413412157 in words is "one trillion, one hundred twenty-three billion, four hundred thirteen million, four hundred twelve thousand, one hundred fifty-seven".