Base | Representation |
---|---|
bin | 11101100100001111111010… |
… | …110110101101001110001101 |
3 | 122001102011100222102110002110 |
4 | 131210033322312231032031 |
5 | 114020440313421001110 |
6 | 1140320541443014233 |
7 | 36250446153000255 |
oct | 3544177266551615 |
9 | 561364328373073 |
10 | 130034343531405 |
11 | 38484289a31735 |
12 | 1270163a66a379 |
13 | 577325579b7c1 |
14 | 24179a5422965 |
15 | 10077596a4a20 |
hex | 7643fadad38d |
130034343531405 has 32 divisors (see below), whose sum is σ = 213702244389504. Its totient is φ = 67469642654976.
The previous prime is 130034343531359. The next prime is 130034343531457. The reversal of 130034343531405 is 504135343430031.
It is not a de Polignac number, because 130034343531405 - 222 = 130034339337101 is a prime.
It is a congruent number.
It is an unprimeable number.
It is a polite number, since it can be written in 31 ways as a sum of consecutive naturals, for example, 8352007 + ... + 18161076.
It is an arithmetic number, because the mean of its divisors is an integer number (6678195137172).
Almost surely, 2130034343531405 is an apocalyptic number.
130034343531405 is a gapful number since it is divisible by the number (15) formed by its first and last digit.
It is an amenable number.
130034343531405 is a deficient number, since it is larger than the sum of its proper divisors (83667900858099).
130034343531405 is a wasteful number, since it uses less digits than its factorization.
130034343531405 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 26521965.
The product of its (nonzero) digits is 388800, while the sum is 39.
Adding to 130034343531405 its reverse (504135343430031), we get a palindrome (634169686961436).
The spelling of 130034343531405 in words is "one hundred thirty trillion, thirty-four billion, three hundred forty-three million, five hundred thirty-one thousand, four hundred five".
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