Base | Representation |
---|---|
bin | 10110010011000101011… |
… | …100110001100110100111 |
3 | 12102111012002022200012002 |
4 | 112103011130301212213 |
5 | 200101143134342020 |
6 | 3131534421115515 |
7 | 215464225461434 |
oct | 26230534614647 |
9 | 5374162280162 |
10 | 1532321012135 |
11 | 54094329a171 |
12 | 208b83258b9b |
13 | b16605cc864 |
14 | 54244052b8b |
15 | 29cd4b45975 |
hex | 164c57319a7 |
1532321012135 has 4 divisors (see below), whose sum is σ = 1838785214568. Its totient is φ = 1225856809704.
The previous prime is 1532321012119. The next prime is 1532321012191. The reversal of 1532321012135 is 5312101232351.
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 1532321012135 - 24 = 1532321012119 is a prime.
It is a Duffinian number.
It is a junction number, because it is equal to n+sod(n) for n = 1532321012098 and 1532321012107.
It is a congruent number.
It is an unprimeable number.
It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 153232101209 + ... + 153232101218.
It is an arithmetic number, because the mean of its divisors is an integer number (459696303642).
Almost surely, 21532321012135 is an apocalyptic number.
1532321012135 is a deficient number, since it is larger than the sum of its proper divisors (306464202433).
1532321012135 is an equidigital number, since it uses as much as digits as its factorization.
1532321012135 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 306464202432.
The product of its (nonzero) digits is 5400, while the sum is 29.
Adding to 1532321012135 its reverse (5312101232351), we get a palindrome (6844422244486).
The spelling of 1532321012135 in words is "one trillion, five hundred thirty-two billion, three hundred twenty-one million, twelve thousand, one hundred thirty-five".
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