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1134333322055 = 5226866664411
BaseRepresentation
bin10000100000011011100…
…010001110001101000111
311000102220122021020110022
4100200123202032031013
5122041103312301210
62225034443445355
7144644560102253
oct20403342161507
94012818236408
101134333322055
113a808265201a
12163a11b2685b
1382c75b5895b
143cc8b231263
151e78ec42155
hex1081b88e347

1134333322055 has 4 divisors (see below), whose sum is σ = 1361199986472. Its totient is φ = 907466657640.

The previous prime is 1134333322049. The next prime is 1134333322069. The reversal of 1134333322055 is 5502233334311.

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

It is not a de Polignac number, because 1134333322055 - 232 = 1130038354759 is a prime.

It is a Duffinian number.

It is a congruent 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 3 ways as a sum of consecutive naturals, for example, 113433332201 + ... + 113433332210.

It is an arithmetic number, because the mean of its divisors is an integer number (340299996618).

Almost surely, 21134333322055 is an apocalyptic number.

1134333322055 is a deficient number, since it is larger than the sum of its proper divisors (226866664417).

1134333322055 is an equidigital number, since it uses as much as digits as its factorization.

1134333322055 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 226866664416.

The product of its (nonzero) digits is 97200, while the sum is 35.

Adding to 1134333322055 its reverse (5502233334311), we get a palindrome (6636566656366).

The spelling of 1134333322055 in words is "one trillion, one hundred thirty-four billion, three hundred thirty-three million, three hundred twenty-two thousand, fifty-five".

Divisors: 1 5 226866664411 1134333322055