Search a number
-
+
3331421011 = 4181254171
BaseRepresentation
bin1100011010010001…
…0111001101010011
322121011122201111101
43012210113031103
523310320433021
61310323525231
7145361416462
oct30644271523
98534581441
103331421011
11145a555a06
1278b829817
13411264049
142386389d9
1514770ca91
hexc6917353

3331421011 has 4 divisors (see below), whose sum is σ = 3412675224. Its totient is φ = 3250166800.

The previous prime is 3331421003. The next prime is 3331421047. The reversal of 3331421011 is 1101241333.

3331421011 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.

It is a semiprime because it is the product of two primes, and also an emirpimes, since its reverse is a distinct semiprime: 1101241333 = 4632378491.

It is a cyclic number.

It is not a de Polignac number, because 3331421011 - 23 = 3331421003 is a prime.

It is a super-2 number, since 2×33314210112 = 22196731905064524242, which contains 22 as substring.

It is a Duffinian number.

It is a self number, because there is not a number n which added to its sum of digits gives 3331421011.

It is not an unprimeable number, because it can be changed into a prime (3331421311) by changing a digit.

It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 40627045 + ... + 40627126.

It is an arithmetic number, because the mean of its divisors is an integer number (853168806).

Almost surely, 23331421011 is an apocalyptic number.

3331421011 is a deficient number, since it is larger than the sum of its proper divisors (81254213).

3331421011 is an equidigital number, since it uses as much as digits as its factorization.

3331421011 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 81254212.

The product of its (nonzero) digits is 216, while the sum is 19.

The square root of 3331421011 is about 57718.4633457960. The cubic root of 3331421011 is about 1493.5158645276.

Adding to 3331421011 its reverse (1101241333), we get a palindrome (4432662344).

It can be divided in two parts, 333142 and 1011, that added together give a triangular number (334153 = T817).

The spelling of 3331421011 in words is "three billion, three hundred thirty-one million, four hundred twenty-one thousand, eleven".

Divisors: 1 41 81254171 3331421011