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132123140111 is a prime number
BaseRepresentation
bin111101100001100100…
…1110010000000001111
3110122000212222221022202
41323003021302000033
54131042010440421
6140410241232115
712355064106134
oct1730311620017
9418025887282
10132123140111
1151040082786
122173393263b
13c5c7a05161
14657545a58b
15368444c10b
hex1ec327200f

132123140111 has 2 divisors, whose sum is σ = 132123140112. Its totient is φ = 132123140110.

The previous prime is 132123140011. The next prime is 132123140113. The reversal of 132123140111 is 111041321231.

It is a strong prime.

It is an emirp because it is prime and its reverse (111041321231) is a distict prime.

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2k-132123140111 is a prime.

Together with 132123140113, it forms a pair of twin primes.

It is a Chen prime.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (132123140113) 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, 66061570055 + 66061570056.

It is an arithmetic number, because the mean of its divisors is an integer number (66061570056).

Almost surely, 2132123140111 is an apocalyptic number.

132123140111 is a deficient number, since it is larger than the sum of its proper divisors (1).

132123140111 is an equidigital number, since it uses as much as digits as its factorization.

132123140111 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 144, while the sum is 20.

Adding to 132123140111 its reverse (111041321231), we get a palindrome (243164461342).

The spelling of 132123140111 in words is "one hundred thirty-two billion, one hundred twenty-three million, one hundred forty thousand, one hundred eleven".