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111130041891749 is a prime number
BaseRepresentation
bin11001010001001001111010…
…101010111011101110100101
3112120110221001200201202220012
4121101021322222323232211
5104031223311211013444
61032204234135220005
732256611015230406
oct3121117252735645
9476427050652805
10111130041891749
1132455a935480a1
12105698b6217605
134a0169701b42a
141d62a192399ad
15ccab32814b9e
hex65127aabbba5

111130041891749 has 2 divisors, whose sum is σ = 111130041891750. Its totient is φ = 111130041891748.

The previous prime is 111130041891719. The next prime is 111130041891763. The reversal of 111130041891749 is 947198140031111.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 59029334133025 + 52100707758724 = 7683055^2 + 7218082^2 .

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2k-111130041891749 is a prime.

It is a junction number, because it is equal to n+sod(n) for n = 111130041891697 and 111130041891706.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (111130041891719) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 55565020945874 + 55565020945875.

It is an arithmetic number, because the mean of its divisors is an integer number (55565020945875).

Almost surely, 2111130041891749 is an apocalyptic number.

It is an amenable number.

111130041891749 is a deficient number, since it is larger than the sum of its proper divisors (1).

111130041891749 is an equidigital number, since it uses as much as digits as its factorization.

111130041891749 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 217728, while the sum is 50.

The spelling of 111130041891749 in words is "one hundred eleven trillion, one hundred thirty billion, forty-one million, eight hundred ninety-one thousand, seven hundred forty-nine".