110100113 has 2 divisors, whose sum is σ = 110100114. Its totient is φ = 110100112.

The previous prime is 110100097. The next prime is 110100121. The reversal of 110100113 is 311001011.

It is an a-pointer prime, because the next prime (110100121) can be obtained adding 110100113 to its sum of digits (8).

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 66961489 + 43138624 = 8183^2 + 6568^2 .

It is an emirp because it is prime and its reverse (311001011) is a distict prime.

It is a cyclic number.

It is not a de Polignac number, because 110100113 - 2⁴ = 110100097 is a prime.

It is a super-2 number, since 2×110100113² = 24244069765225538, which contains 22 as substring.

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 110100113.

It is not a weakly prime, because it can be changed into another prime (110100013) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (17) of ones.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 55050056 + 55050057.

It is an arithmetic number, because the mean of its divisors is an integer number (55050057).

Almost surely, 2^110100113 is an apocalyptic number.

It is an amenable number.

110100113 is a deficient number, since it is larger than the sum of its proper divisors (1).

110100113 is an equidigital number, since it uses as much as digits as its factorization.

110100113 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 3, while the sum is 8.

The square root of 110100113 is about 10492.8600962750. The cubic root of 110100113 is about 479.2873002417.

Adding to 110100113 its reverse (311001011), we get a palindrome (421101124).

The spelling of 110100113 in words is "one hundred ten million, one hundred thousand, one hundred thirteen".