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14051113555 = 54787587053
BaseRepresentation
bin11010001011000001…
…01111101001010011
31100021020122112012011
431011200233221103
5212234041113210
610242140043351
71005125331433
oct150540575123
940236575164
1014051113555
115a6053a646
122881833557
13142c09709b
149741c02c3
1557387cd8a
hex34582fa53

14051113555 has 8 divisors (see below), whose sum is σ = 16864887312. Its totient is φ = 11238523488.

The previous prime is 14051113519. The next prime is 14051113607. The reversal of 14051113555 is 55531115041.

It is a happy number.

14051113555 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.

It is a sphenic number, since it is the product of 3 distinct primes.

It is a cyclic number.

It is not a de Polignac number, because 14051113555 - 217 = 14050982483 is a prime.

It is a Duffinian number.

It is an unprimeable number.

It is a pernicious number, because its binary representation contains a prime number (17) of ones.

It is a polite number, since it can be written in 7 ways as a sum of consecutive naturals, for example, 269592 + ... + 317461.

It is an arithmetic number, because the mean of its divisors is an integer number (2108110914).

Almost surely, 214051113555 is an apocalyptic number.

14051113555 is a deficient number, since it is larger than the sum of its proper divisors (2813773757).

14051113555 is an equidigital number, since it uses as much as digits as its factorization.

14051113555 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 591845.

The product of its (nonzero) digits is 7500, while the sum is 31.

Adding to 14051113555 its reverse (55531115041), we get a palindrome (69582228596).

The spelling of 14051113555 in words is "fourteen billion, fifty-one million, one hundred thirteen thousand, five hundred fifty-five".

Divisors: 1 5 4787 23935 587053 2935265 2810222711 14051113555