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20010311133 = 36670103711
BaseRepresentation
bin10010101000101101…
…010001110111011101
31220122112221010110120
4102220231101313131
5311440114424013
613105334225153
71305605560341
oct225055216735
956575833416
1020010311133
118539331a1a
123a654b51b9
131b6b877cb7
14d7b80b621
157c1b06123
hex4a8b51ddd

20010311133 has 4 divisors (see below), whose sum is σ = 26680414848. Its totient is φ = 13340207420.

The previous prime is 20010311081. The next prime is 20010311143. The reversal of 20010311133 is 33111301002.

It is a semiprime because it is the product of two primes, and also a Blum integer, because the two primes are equal to 3 mod 4.

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2k-20010311133 is a prime.

It is a congruent number.

It is not an unprimeable number, because it can be changed into a prime (20010311143) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (19) of ones.

It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 3335051853 + ... + 3335051858.

It is an arithmetic number, because the mean of its divisors is an integer number (6670103712).

Almost surely, 220010311133 is an apocalyptic number.

It is an amenable number.

20010311133 is a deficient number, since it is larger than the sum of its proper divisors (6670103715).

20010311133 is an equidigital number, since it uses as much as digits as its factorization.

20010311133 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 6670103714.

The product of its (nonzero) digits is 54, while the sum is 15.

Adding to 20010311133 its reverse (33111301002), we get a palindrome (53121612135).

The spelling of 20010311133 in words is "twenty billion, ten million, three hundred eleven thousand, one hundred thirty-three".

Divisors: 1 3 6670103711 20010311133