1100030012117 has 2 divisors, whose sum is σ = 1100030012118. Its totient is φ = 1100030012116.

The previous prime is 1100030012093. The next prime is 1100030012119. The reversal of 1100030012117 is 7112100300011.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 961358679121 + 138671332996 = 980489^2 + 372386^2 .

It is a cyclic number.

It is not a de Polignac number, because 1100030012117 - 2¹⁴ = 1100029995733 is a prime.

Together with 1100030012119, it forms a pair of twin primes.

It is a Chen prime.

It is a junction number, because it is equal to n+sod(n) for n = 1100030012095 and 1100030012104.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (1100030012119) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 550015006058 + 550015006059.

It is an arithmetic number, because the mean of its divisors is an integer number (550015006059).

Almost surely, 2^{1100030012117} is an apocalyptic number.

It is an amenable number.

1100030012117 is a deficient number, since it is larger than the sum of its proper divisors (1).

1100030012117 is an equidigital number, since it uses as much as digits as its factorization.

1100030012117 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 42, while the sum is 17.

Adding to 1100030012117 its reverse (7112100300011), we get a palindrome (8212130312128).

The spelling of 1100030012117 in words is "one trillion, one hundred billion, thirty million, twelve thousand, one hundred seventeen".