100043311 has 2 divisors, whose sum is σ = 100043312. Its totient is φ = 100043310.

The previous prime is 100043299. The next prime is 100043347. The reversal of 100043311 is 113340001.

It is a weak prime.

It is an emirp because it is prime and its reverse (113340001) is a distict prime.

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2^k-100043311 is a prime.

It is a Chen prime.

It is a junction number, because it is equal to n+sod(n) for n = 100043291 and 100043300.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (100043371) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 50021655 + 50021656.

It is an arithmetic number, because the mean of its divisors is an integer number (50021656).

Almost surely, 2^100043311 is an apocalyptic number.

100043311 is a deficient number, since it is larger than the sum of its proper divisors (1).

100043311 is an equidigital number, since it uses as much as digits as its factorization.

100043311 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 36, while the sum is 13.

The square root of 100043311 is about 10002.1653155704. The cubic root of 100043311 is about 464.2258843073.

Adding to 100043311 its reverse (113340001), we get a palindrome (213383312).

The spelling of 100043311 in words is "one hundred million, forty-three thousand, three hundred eleven".