Base | Representation |
---|---|
bin | 111100100011011111… |
… | …1110001101011011011 |
3 | 110102122201110200212010 |
4 | 1321012333301223123 |
5 | 4112310234030210 |
6 | 135423431321003 |
7 | 12252341655021 |
oct | 1710677615333 |
9 | 412581420763 |
10 | 130040142555 |
11 | 501712a78a4 |
12 | 21252220163 |
13 | c3552bb21b |
14 | 6418958711 |
15 | 35b1642920 |
hex | 1e46ff1adb |
130040142555 has 32 divisors (see below), whose sum is σ = 208397952000. Its totient is φ = 69243609600.
The previous prime is 130040142511. The next prime is 130040142559. The reversal of 130040142555 is 555241040031.
It is not a de Polignac number, because 130040142555 - 214 = 130040126171 is a prime.
It is not an unprimeable number, because it can be changed into a prime (130040142559) 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, 37163196 + ... + 37166694.
It is an arithmetic number, because the mean of its divisors is an integer number (6512436000).
Almost surely, 2130040142555 is an apocalyptic number.
130040142555 is a gapful number since it is divisible by the number (15) formed by its first and last digit.
130040142555 is a deficient number, since it is larger than the sum of its proper divisors (78357809445).
130040142555 is a wasteful number, since it uses less digits than its factorization.
130040142555 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 6771.
The product of its (nonzero) digits is 12000, while the sum is 30.
Adding to 130040142555 its reverse (555241040031), we get a palindrome (685281182586).
The spelling of 130040142555 in words is "one hundred thirty billion, forty million, one hundred forty-two thousand, five hundred fifty-five".
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