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11028490909403 = 111002590082673
BaseRepresentation
bin1010000001111100010101…
…0010000111011011011011
31110001022110111022211120012
42200133011102013123123
52421142322133100103
635242230242352135
72215532200003202
oct240370522073333
943038414284505
1011028490909403
11357217a673930
1212a148b35b64b
1361cc9cc63226
142a1ad33d2239
15141d22a902d8
hexa07c54876db

11028490909403 has 4 divisors (see below), whose sum is σ = 12031080992088. Its totient is φ = 10025900826720.

The previous prime is 11028490909399. The next prime is 11028490909421. The reversal of 11028490909403 is 30490909482011.

11028490909403 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.

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 11028490909403 - 22 = 11028490909399 is a prime.

It is a super-2 number, since 2×110284909094032 (a number of 27 digits) contains 22 as substring.

It is a Duffinian number.

It is not an unprimeable number, because it can be changed into a prime (11028490909453) by changing a digit.

It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 501295041326 + ... + 501295041347.

It is an arithmetic number, because the mean of its divisors is an integer number (3007770248022).

Almost surely, 211028490909403 is an apocalyptic number.

11028490909403 is a deficient number, since it is larger than the sum of its proper divisors (1002590082685).

11028490909403 is a wasteful number, since it uses less digits than its factorization.

11028490909403 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 1002590082684.

The product of its (nonzero) digits is 559872, while the sum is 50.

The spelling of 11028490909403 in words is "eleven trillion, twenty-eight billion, four hundred ninety million, nine hundred nine thousand, four hundred three".

Divisors: 1 11 1002590082673 11028490909403