1100100999953 has 2 divisors, whose sum is σ = 1100100999954. Its totient is φ = 1100100999952.

The previous prime is 1100100999947. The next prime is 1100101000051. The reversal of 1100100999953 is 3599990010011.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 964111899664 + 135989100289 = 981892^2 + 368767^2 .

It is an emirp because it is prime and its reverse (3599990010011) is a distict prime.

It is a cyclic number.

It is not a de Polignac number, because 1100100999953 - 2⁶ = 1100100999889 is a prime.

It is a Sophie Germain prime.

It is a Curzon number.

It is a junction number, because it is equal to n+sod(n) for n = 1100100999898 and 1100100999907.

It is not a weakly prime, because it can be changed into another prime (1100100909953) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 550050499976 + 550050499977.

It is an arithmetic number, because the mean of its divisors is an integer number (550050499977).

Almost surely, 2^{1100100999953} is an apocalyptic number.

It is an amenable number.

1100100999953 is a deficient number, since it is larger than the sum of its proper divisors (1).

1100100999953 is an equidigital number, since it uses as much as digits as its factorization.

1100100999953 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 98415, while the sum is 47.

The spelling of 1100100999953 in words is "one trillion, one hundred billion, one hundred million, nine hundred ninety-nine thousand, nine hundred fifty-three".