Base | Representation |
---|---|
bin | 10011100100000000100… |
… | …100000010111110111011 |
3 | 11202111221221222211021221 |
4 | 103210000210002332323 |
5 | 134011144024304201 |
6 | 2505324551412511 |
7 | 166060566162004 |
oct | 23440044027673 |
9 | 4674857884257 |
10 | 1344334213051 |
11 | 479146686695 |
12 | 19865a984137 |
13 | 99a020b79bb |
14 | 490cd6281ab |
15 | 24e8119d4a1 |
hex | 13900902fbb |
1344334213051 has 2 divisors, whose sum is σ = 1344334213052. Its totient is φ = 1344334213050.
The previous prime is 1344334212983. The next prime is 1344334213109. The reversal of 1344334213051 is 1503124334431.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 1344334213051 - 215 = 1344334180283 is a prime.
It is a junction number, because it is equal to n+sod(n) for n = 1344334212998 and 1344334213016.
It is not a weakly prime, because it can be changed into another prime (1344334213151) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 672167106525 + 672167106526.
It is an arithmetic number, because the mean of its divisors is an integer number (672167106526).
Almost surely, 21344334213051 is an apocalyptic number.
1344334213051 is a deficient number, since it is larger than the sum of its proper divisors (1).
1344334213051 is an equidigital number, since it uses as much as digits as its factorization.
1344334213051 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 51840, while the sum is 34.
Adding to 1344334213051 its reverse (1503124334431), we get a palindrome (2847458547482).
The spelling of 1344334213051 in words is "one trillion, three hundred forty-four billion, three hundred thirty-four million, two hundred thirteen thousand, fifty-one".
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