Search a number
-
+
2013113 is a prime number
BaseRepresentation
bin111101011011110111001
310210021110202
413223132321
51003404423
6111051545
723053064
oct7533671
93707422
102013113
111155533
12810bb5
135563bb
143a58db
1529b728
hex1eb7b9

2013113 has 2 divisors, whose sum is σ = 2013114. Its totient is φ = 2013112.

The previous prime is 2013101. The next prime is 2013119. The reversal of 2013113 is 3113102.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 1951609 + 61504 = 1397^2 + 248^2 .

It is a cyclic number.

It is not a de Polignac number, because 2013113 - 26 = 2013049 is a prime.

It is a Sophie Germain prime.

It is a Curzon number.

It is a zygodrome in base 11.

It is a junction number, because it is equal to n+sod(n) for n = 2013094 and 2013103.

It is not a weakly prime, because it can be changed into another prime (2013119) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 1006556 + 1006557.

It is an arithmetic number, because the mean of its divisors is an integer number (1006557).

22013113 is an apocalyptic number.

It is an amenable number.

2013113 is a deficient number, since it is larger than the sum of its proper divisors (1).

2013113 is an equidigital number, since it uses as much as digits as its factorization.

2013113 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 18, while the sum is 11.

The square root of 2013113 is about 1418.8421335723. The cubic root of 2013113 is about 126.2668611274.

Subtracting from 2013113 its sum of digits (11), we obtain a palindrome (2013102).

Adding to 2013113 its reverse (3113102), we get a palindrome (5126215).

The spelling of 2013113 in words is "two million, thirteen thousand, one hundred thirteen".