Base | Representation |
---|---|
bin | 11001000001011101… |
… | …01100000101110100 |
3 | 1021200020110112000202 |
4 | 30200232230011310 |
5 | 210003101341040 |
6 | 10101013300032 |
7 | 653623134542 |
oct | 144056540564 |
9 | 37606415022 |
10 | 13434012020 |
11 | 5774171080 |
12 | 272b034618 |
13 | 1361291450 |
14 | 916264d92 |
15 | 5395d3015 |
hex | 320bac174 |
13434012020 has 96 divisors (see below), whose sum is σ = 33472365696. Its totient is φ = 4464576000.
The previous prime is 13434012019. The next prime is 13434012029. The reversal of 13434012020 is 2021043431.
It is a Harshad number since it is a multiple of its sum of digits (20).
It is a junction number, because it is equal to n+sod(n) for n = 13434011983 and 13434012001.
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (13434012029) by changing a digit.
It is a polite number, since it can be written in 31 ways as a sum of consecutive naturals, for example, 265607 + ... + 312113.
It is an arithmetic number, because the mean of its divisors is an integer number (348670476).
Almost surely, 213434012020 is an apocalyptic number.
13434012020 is a gapful number since it is divisible by the number (10) formed by its first and last digit.
It is an amenable number.
It is a practical number, because each smaller number is the sum of distinct divisors of 13434012020, and also a Zumkeller number, because its divisors can be partitioned in two sets with the same sum (16736182848).
13434012020 is an abundant number, since it is smaller than the sum of its proper divisors (20038353676).
It is a pseudoperfect number, because it is the sum of a subset of its proper divisors.
13434012020 is a wasteful number, since it uses less digits than its factorization.
13434012020 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 46641 (or 46639 counting only the distinct ones).
The product of its (nonzero) digits is 576, while the sum is 20.
Adding to 13434012020 its reverse (2021043431), we get a palindrome (15455055451).
The spelling of 13434012020 in words is "thirteen billion, four hundred thirty-four million, twelve thousand, twenty".
• e-mail: info -at- numbersaplenty.com • Privacy notice • done in 0.075 sec. • engine limits •