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130031031011200 = 2752116595605559
BaseRepresentation
bin11101100100001100110101…
…011010011100011110000000
3122001101211211010022222121221
4131210030311122130132000
5114020412022414324300
61140315233032124424
736250301111021044
oct3544146532343600
9561354733288557
10130031031011200
113848293a116230
1212700a75236714
135772b48424787
14241776d508024
15100761397c31a
hex76433569c780

130031031011200 has 192 divisors, whose sum is σ = 350950658256000. Its totient is φ = 47212251699200.

The previous prime is 130031031011197. The next prime is 130031031011201. The reversal of 130031031011200 is 2110130130031.

It is a super-4 number, since 4×1300310310112004 (a number of 58 digits) contains 4444 as substring.

It is a Harshad number since it is a multiple of its sum of digits (16).

It is a congruent number.

It is not an unprimeable number, because it can be changed into a prime (130031031011201) by changing a digit.

It is a polite number, since it can be written in 23 ways as a sum of consecutive naturals, for example, 20394021 + ... + 25999579.

It is an arithmetic number, because the mean of its divisors is an integer number (1827868011750).

Almost surely, 2130031031011200 is an apocalyptic number.

130031031011200 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 130031031011200, and also a Zumkeller number, because its divisors can be partitioned in two sets with the same sum (175475329128000).

130031031011200 is an abundant number, since it is smaller than the sum of its proper divisors (220919627244800).

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

130031031011200 is a wasteful number, since it uses less digits than its factorization.

130031031011200 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 5606253 (or 5606236 counting only the distinct ones).

The product of its (nonzero) digits is 54, while the sum is 16.

Adding to 130031031011200 its reverse (2110130130031), we get a palindrome (132141161141231).

The spelling of 130031031011200 in words is "one hundred thirty trillion, thirty-one billion, thirty-one million, eleven thousand, two hundred".