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19813 is a prime number
BaseRepresentation
bin100110101100101
31000011211
410311211
51113223
6231421
7111523
oct46545
930154
1019813
1113982
12b571
139031
147313
155d0d
hex4d65

19813 has 2 divisors, whose sum is σ = 19814. Its totient is φ = 19812.

The previous prime is 19801. The next prime is 19819. The reversal of 19813 is 31891.

Subtracting from 19813 its sum of digits (22), we obtain a palindrome (19791).

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 10404 + 9409 = 102^2 + 97^2 .

It is an emirp because it is prime and its reverse (31891) is a distict prime.

It is a cyclic number.

It is not a de Polignac number, because 19813 - 29 = 19301 is a prime.

It is a super-3 number, since 3×198133 = 23333074802391, which contains 333 as substring.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (19819) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 9906 + 9907.

It is an arithmetic number, because the mean of its divisors is an integer number (9907).

219813 is an apocalyptic number.

It is an amenable number.

19813 is a deficient number, since it is larger than the sum of its proper divisors (1).

19813 is an equidigital number, since it uses as much as digits as its factorization.

19813 is an evil number, because the sum of its binary digits is even.

The product of its digits is 216, while the sum is 22.

The square root of 19813 is about 140.7586587035. The cubic root of 19813 is about 27.0593117707.

The spelling of 19813 in words is "nineteen thousand, eight hundred thirteen".