Base | Representation |
---|---|
bin | 10001101010110100100… |
… | …100111000000110110011 |
3 | 11022002002101222100101022 |
4 | 101222310210320012303 |
5 | 124343200321441342 |
6 | 2325444525330055 |
7 | 153503156425331 |
oct | 21526444700663 |
9 | 4262071870338 |
10 | 1214210343347 |
11 | 428a420a2401 |
12 | 1773a470692b |
13 | 8a665662ab1 |
14 | 42aa7916b51 |
15 | 218b751c9d2 |
hex | 11ab49381b3 |
1214210343347 has 2 divisors, whose sum is σ = 1214210343348. Its totient is φ = 1214210343346.
The previous prime is 1214210343257. The next prime is 1214210343361. The reversal of 1214210343347 is 7433430124121.
It is a strong prime.
It is an emirp because it is prime and its reverse (7433430124121) is a distict prime.
It is a cyclic number.
It is not a de Polignac number, because 1214210343347 - 214 = 1214210326963 is a prime.
It is not a weakly prime, because it can be changed into another prime (1214210343847) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (19) of ones.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 607105171673 + 607105171674.
It is an arithmetic number, because the mean of its divisors is an integer number (607105171674).
Almost surely, 21214210343347 is an apocalyptic number.
1214210343347 is a deficient number, since it is larger than the sum of its proper divisors (1).
1214210343347 is an equidigital number, since it uses as much as digits as its factorization.
1214210343347 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 48384, while the sum is 35.
Adding to 1214210343347 its reverse (7433430124121), we get a palindrome (8647640467468).
The spelling of 1214210343347 in words is "one trillion, two hundred fourteen billion, two hundred ten million, three hundred forty-three thousand, three hundred forty-seven".
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