Base | Representation |
---|---|
bin | 10100110101110110100… |
… | …011111001001101000111 |
3 | 12001220210101212110012112 |
4 | 110311312203321031013 |
5 | 141431133041333112 |
6 | 3013541044011235 |
7 | 205321402410461 |
oct | 24656643711507 |
9 | 5056711773175 |
10 | 1432213230407 |
11 | 502442038063 |
12 | 1b16a5390b1b |
13 | a509868365b |
14 | 4d468977131 |
15 | 273c621aa22 |
hex | 14d768f9347 |
1432213230407 has 2 divisors, whose sum is σ = 1432213230408. Its totient is φ = 1432213230406.
The previous prime is 1432213230379. The next prime is 1432213230421. The reversal of 1432213230407 is 7040323122341.
It is a strong prime.
It is an emirp because it is prime and its reverse (7040323122341) is a distict prime.
It is a cyclic number.
It is not a de Polignac number, because 1432213230407 - 216 = 1432213164871 is a prime.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (1432213239407) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (23) of ones.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 716106615203 + 716106615204.
It is an arithmetic number, because the mean of its divisors is an integer number (716106615204).
Almost surely, 21432213230407 is an apocalyptic number.
1432213230407 is a deficient number, since it is larger than the sum of its proper divisors (1).
1432213230407 is an equidigital number, since it uses as much as digits as its factorization.
1432213230407 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 24192, while the sum is 32.
Adding to 1432213230407 its reverse (7040323122341), we get a palindrome (8472536352748).
The spelling of 1432213230407 in words is "one trillion, four hundred thirty-two billion, two hundred thirteen million, two hundred thirty thousand, four hundred seven".
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