Base | Representation |
---|---|
bin | 100111011000011… |
… | …0010100111101110 |
3 | 10102002110202000000 |
4 | 1032300302213232 |
5 | 10201240204402 |
6 | 335042245130 |
7 | 44513553222 |
oct | 11660624756 |
9 | 3362422000 |
10 | 1321413102 |
11 | 6189a3249 |
12 | 30a6567a6 |
13 | 1809c4659 |
14 | c76d6782 |
15 | 7b01ec1c |
hex | 4ec329ee |
1321413102 has 56 divisors (see below), whose sum is σ = 3128297160. Its totient is φ = 417279600.
The previous prime is 1321413097. The next prime is 1321413103. The reversal of 1321413102 is 2013141231.
It is a super-3 number, since 3×13214131023 (a number of 28 digits) contains 333 as substring. Note that it is a super-d number also for d = 2.
It is a Harshad number since it is a multiple of its sum of digits (18).
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (1321413103) 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 27 ways as a sum of consecutive naturals, for example, 3852 + ... + 51552.
Almost surely, 21321413102 is an apocalyptic number.
It is a practical number, because each smaller number is the sum of distinct divisors of 1321413102, and also a Zumkeller number, because its divisors can be partitioned in two sets with the same sum (1564148580).
1321413102 is an abundant number, since it is smaller than the sum of its proper divisors (1806884058).
It is a pseudoperfect number, because it is the sum of a subset of its proper divisors.
1321413102 is an equidigital number, since it uses as much as digits as its factorization.
1321413102 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 47740 (or 47725 counting only the distinct ones).
The product of its (nonzero) digits is 144, while the sum is 18.
The square root of 1321413102 is about 36351.2462234791. The cubic root of 1321413102 is about 1097.3526148805.
Adding to 1321413102 its reverse (2013141231), we get a palindrome (3334554333).
The spelling of 1321413102 in words is "one billion, three hundred twenty-one million, four hundred thirteen thousand, one hundred two".
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