Base | Representation |
---|---|
bin | 1100000010111010111100… |
… | …1110100100001100100011 |
3 | 1201220010220201012221112111 |
4 | 3000232233032210030203 |
5 | 3213443333421122033 |
6 | 44100205231540151 |
7 | 2534604460360624 |
oct | 300565716441443 |
9 | 51803821187474 |
10 | 13244324004643 |
11 | 4246982692696 |
12 | 159aa0938a657 |
13 | 750c1bb32474 |
14 | 33b058a89a4b |
15 | 17e7ae0197cd |
hex | c0baf3a4323 |
13244324004643 has 2 divisors, whose sum is σ = 13244324004644. Its totient is φ = 13244324004642.
The previous prime is 13244324004631. The next prime is 13244324004647. The reversal of 13244324004643 is 34640042344231.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 13244324004643 - 225 = 13244290450211 is a prime.
It is a junction number, because it is equal to n+sod(n) for n = 13244324004596 and 13244324004605.
It is not a weakly prime, because it can be changed into another prime (13244324004647) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6622162002321 + 6622162002322.
It is an arithmetic number, because the mean of its divisors is an integer number (6622162002322).
Almost surely, 213244324004643 is an apocalyptic number.
13244324004643 is a deficient number, since it is larger than the sum of its proper divisors (1).
13244324004643 is an equidigital number, since it uses as much as digits as its factorization.
13244324004643 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 663552, while the sum is 40.
Adding to 13244324004643 its reverse (34640042344231), we get a palindrome (47884366348874).
The spelling of 13244324004643 in words is "thirteen trillion, two hundred forty-four billion, three hundred twenty-four million, four thousand, six hundred forty-three".
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