Base | Representation |
---|---|
bin | 1010000111000011100100… |
… | …1110010011100011110011 |
3 | 1110100201012112021000011121 |
4 | 2201300321032103203303 |
5 | 2424112223331212431 |
6 | 35350435103123111 |
7 | 2225062103655130 |
oct | 241607116234363 |
9 | 43321175230147 |
10 | 11116335413491 |
11 | 35a64585a1a40 |
12 | 12b65062a3497 |
13 | 62835c336829 |
14 | 2a6067d19587 |
15 | 144264a91611 |
hex | a1c393938f3 |
11116335413491 has 16 divisors (see below), whose sum is σ = 13860023505408. Its totient is φ = 8661644395200.
The previous prime is 11116335413479. The next prime is 11116335413539. The reversal of 11116335413491 is 19431453361111.
11116335413491 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.
It is a cyclic number.
It is not a de Polignac number, because 11116335413491 - 29 = 11116335412979 is a prime.
It is a Duffinian number.
It is not an unprimeable number, because it can be changed into a prime (11116335413441) by changing a digit.
It is a polite number, since it can be written in 15 ways as a sum of consecutive naturals, for example, 2079255 + ... + 5153248.
It is an arithmetic number, because the mean of its divisors is an integer number (866251469088).
Almost surely, 211116335413491 is an apocalyptic number.
11116335413491 is a gapful number since it is divisible by the number (11) formed by its first and last digit.
11116335413491 is a deficient number, since it is larger than the sum of its proper divisors (2743688091917).
11116335413491 is a wasteful number, since it uses less digits than its factorization.
11116335413491 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 7252482.
The product of its digits is 116640, while the sum is 43.
The spelling of 11116335413491 in words is "eleven trillion, one hundred sixteen billion, three hundred thirty-five million, four hundred thirteen thousand, four hundred ninety-one".
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