Base | Representation |
---|---|
bin | 11101010010110111111… |
… | …101110100010100001000 |
3 | 21010110020120101101010200 |
4 | 131102313331310110020 |
5 | 230440343032130440 |
6 | 4140453014412200 |
7 | 265305160161165 |
oct | 35226775642410 |
9 | 7113216341120 |
10 | 2013131130120 |
11 | 70684504a240 |
12 | 2861a9661060 |
13 | 117ab6b59056 |
14 | 6d6167d106c |
15 | 37575bd2c30 |
hex | 1d4b7f74508 |
2013131130120 has 384 divisors, whose sum is σ = 7460549141760. Its totient is φ = 466013306880.
The previous prime is 2013131130113. The next prime is 2013131130203. The reversal of 2013131130120 is 210311313102.
2013131130120 is a `hidden beast` number, since 2 + 0 + 1 + 31 + 311 + 301 + 20 = 666.
It is a Harshad number since it is a multiple of its sum of digits (18).
It is a junction number, because it is equal to n+sod(n) for n = 2013131130093 and 2013131130102.
It is an unprimeable number.
It is a polite number, since it can be written in 95 ways as a sum of consecutive naturals, for example, 53991804 + ... + 54029076.
It is an arithmetic number, because the mean of its divisors is an integer number (19428513390).
Almost surely, 22013131130120 is an apocalyptic number.
2013131130120 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 2013131130120, and also a Zumkeller number, because its divisors can be partitioned in two sets with the same sum (3730274570880).
2013131130120 is an abundant number, since it is smaller than the sum of its proper divisors (5447418011640).
It is a pseudoperfect number, because it is the sum of a subset of its proper divisors.
2013131130120 is a wasteful number, since it uses less digits than its factorization.
2013131130120 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 37917 (or 37910 counting only the distinct ones).
The product of its (nonzero) digits is 108, while the sum is 18.
Adding to 2013131130120 its reverse (210311313102), we get a palindrome (2223442443222).
The spelling of 2013131130120 in words is "two trillion, thirteen billion, one hundred thirty-one million, one hundred thirty thousand, one hundred twenty".
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