Base | Representation |
---|---|
bin | 11110001010011001110110… |
… | …001101110101001011011001 |
3 | 122101200211012122110102020212 |
4 | 132022121312031311023121 |
5 | 114341420142332211213 |
6 | 1150045252140300505 |
7 | 36641050564543544 |
oct | 3612316615651331 |
9 | 571624178412225 |
10 | 132656343241433 |
11 | 392a526a463850 |
12 | 12a65830b41735 |
13 | 590359374036b |
14 | 24a885bc5d85b |
15 | 1050a68ac29a8 |
hex | 78a6763752d9 |
132656343241433 has 16 divisors (see below), whose sum is σ = 144975349469952. Its totient is φ = 120380566914000.
The previous prime is 132656343241417. The next prime is 132656343241447. The reversal of 132656343241433 is 334142343656231.
It is a cyclic number.
It is not a de Polignac number, because 132656343241433 - 24 = 132656343241417 is a prime.
It is a Duffinian number.
It is not an unprimeable number, because it can be changed into a prime (132656343241453) 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, 520152638 + ... + 520407608.
It is an arithmetic number, because the mean of its divisors is an integer number (9060959341872).
Almost surely, 2132656343241433 is an apocalyptic number.
It is an amenable number.
132656343241433 is a deficient number, since it is larger than the sum of its proper divisors (12319006228519).
132656343241433 is a wasteful number, since it uses less digits than its factorization.
132656343241433 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 339556.
The product of its digits is 11197440, while the sum is 50.
Adding to 132656343241433 its reverse (334142343656231), we get a palindrome (466798686897664).
The spelling of 132656343241433 in words is "one hundred thirty-two trillion, six hundred fifty-six billion, three hundred forty-three million, two hundred forty-one thousand, four hundred thirty-three".
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