Base | Representation |
---|---|
bin | 10100100100000100000… |
… | …111100111101010001111 |
3 | 12000002111022001211101101 |
4 | 110210010013213222033 |
5 | 141123024004043303 |
6 | 3001101511123531 |
7 | 204044160155362 |
oct | 24440407475217 |
9 | 5002438054341 |
10 | 1413113346703 |
11 | 4a533070a803 |
12 | 1a9a549735a7 |
13 | a3343613316 |
14 | 4c57603c0d9 |
15 | 26b59500d1d |
hex | 149041e7a8f |
1413113346703 has 2 divisors, whose sum is σ = 1413113346704. Its totient is φ = 1413113346702.
The previous prime is 1413113346661. The next prime is 1413113346713. The reversal of 1413113346703 is 3076433113141.
It is a strong prime.
It is an emirp because it is prime and its reverse (3076433113141) is a distict prime.
It is a cyclic number.
It is not a de Polignac number, because 1413113346703 - 221 = 1413111249551 is a prime.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (1413113346713) 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, 706556673351 + 706556673352.
It is an arithmetic number, because the mean of its divisors is an integer number (706556673352).
Almost surely, 21413113346703 is an apocalyptic number.
1413113346703 is a deficient number, since it is larger than the sum of its proper divisors (1).
1413113346703 is an equidigital number, since it uses as much as digits as its factorization.
1413113346703 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 54432, while the sum is 37.
Adding to 1413113346703 its reverse (3076433113141), we get a palindrome (4489546459844).
The spelling of 1413113346703 in words is "one trillion, four hundred thirteen billion, one hundred thirteen million, three hundred forty-six thousand, seven hundred three".
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