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11143303115 = 57318380089
BaseRepresentation
bin10100110000011000…
…10100111111001011
31001202121002110011012
422120030110333023
5140310141144430
65041423354135
7543066341120
oct123014247713
931677073135
1011143303115
1147a911661a
1221aba5294b
1310878202a9
14779d34747
15453449695
hex298314fcb

11143303115 has 8 divisors (see below), whose sum is σ = 15282244320. Its totient is φ = 7641122112.

The previous prime is 11143303111. The next prime is 11143303133. The reversal of 11143303115 is 51130334111.

11143303115 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 11143303115 - 22 = 11143303111 is a prime.

It is not an unprimeable number, because it can be changed into a prime (11143303111) by changing a digit.

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, 159190010 + ... + 159190079.

It is an arithmetic number, because the mean of its divisors is an integer number (1910280540).

Almost surely, 211143303115 is an apocalyptic number.

11143303115 is a deficient number, since it is larger than the sum of its proper divisors (4138941205).

11143303115 is an equidigital number, since it uses as much as digits as its factorization.

11143303115 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 318380101.

The product of its (nonzero) digits is 540, while the sum is 23.

Adding to 11143303115 its reverse (51130334111), we get a palindrome (62273637226).

The spelling of 11143303115 in words is "eleven billion, one hundred forty-three million, three hundred three thousand, one hundred fifteen".

Divisors: 1 5 7 35 318380089 1591900445 2228660623 11143303115