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20031364235 = 54006272847
BaseRepresentation
bin10010101001111101…
…100101110010001011
31220201000112201220002
4102221331211302023
5312011012123420
613111405353215
71306252533551
oct225175456213
956630481802
1020031364235
11854a201483
123a7056880b
131b7303985c
14d8052bbd1
157c38b4075
hex4a9f65c8b

20031364235 has 4 divisors (see below), whose sum is σ = 24037637088. Its totient is φ = 16025091384.

The previous prime is 20031364207. The next prime is 20031364247. The reversal of 20031364235 is 53246313002.

It is a semiprime because it is the product of two primes.

It is a cyclic number.

It is not a de Polignac number, because 20031364235 - 214 = 20031347851 is a prime.

It is a Duffinian number.

It is a junction number, because it is equal to n+sod(n) for n = 20031364198 and 20031364207.

It is an unprimeable number.

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, 2003136419 + ... + 2003136428.

It is an arithmetic number, because the mean of its divisors is an integer number (6009409272).

Almost surely, 220031364235 is an apocalyptic number.

20031364235 is a deficient number, since it is larger than the sum of its proper divisors (4006272853).

20031364235 is an equidigital number, since it uses as much as digits as its factorization.

20031364235 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 4006272852.

The product of its (nonzero) digits is 12960, while the sum is 29.

Adding to 20031364235 its reverse (53246313002), we get a palindrome (73277677237).

The spelling of 20031364235 in words is "twenty billion, thirty-one million, three hundred sixty-four thousand, two hundred thirty-five".

Divisors: 1 5 4006272847 20031364235