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2013102000000 = 2732567131229
BaseRepresentation
bin11101010010110110001…
…110101100011110000000
321010110011112120102010100
4131102312032230132000
5230440313103000000
64140450054202400
7265304354444550
oct35226616543600
97113145512110
102013102000000
1170682a663363
1228619b953400
13117ab0ac9cb0
146d6129ab160
153757337ba00
hex1d4b63ac780

2013102000000 has 1344 divisors, whose sum is σ = 8919307706400. Its totient is φ = 424396800000.

The previous prime is 2013101999989. The next prime is 2013102000037. The reversal of 2013102000000 is 2013102.

It is a happy number.

It is a tau number, because it is divible by the number of its divisors (1344).

It is a super-4 number, since 4×20131020000004 (a number of 50 digits) contains 4444 as substring.

It is a Harshad number since it is a multiple of its sum of digits (9).

It is a super Niven number, because it is divisible the sum of any subset of its (nonzero) digits.

It is a congruent number.

It is an unprimeable number.

It is a polite number, since it can be written in 167 ways as a sum of consecutive naturals, for example, 1637999386 + ... + 1638000614.

Almost surely, 22013102000000 is an apocalyptic number.

2013102000000 is a gapful number since it is divisible by the number (20) 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 2013102000000, and also a Zumkeller number, because its divisors can be partitioned in two sets with the same sum (4459653853200).

2013102000000 is an abundant number, since it is smaller than the sum of its proper divisors (6906205706400).

It is a pseudoperfect number, because it is the sum of a subset of its proper divisors.

2013102000000 is an equidigital number, since it uses as much as digits as its factorization.

2013102000000 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 1299 (or 1259 counting only the distinct ones).

The product of its (nonzero) digits is 12, while the sum is 9.

Adding to 2013102000000 its reverse (2013102), we get a palindrome (2013104013102).

2013102000000 divided by its reverse (2013102) gives a 6-th power (1000000 = 106).

The spelling of 2013102000000 in words is "two trillion, thirteen billion, one hundred two million", and thus it is an aban number.