Base | Representation |
---|---|
bin | 101110110001100101110… |
… | …010110000010010010001 |
3 | 102101021210010002012002211 |
4 | 232301211302300102101 |
5 | 410130433343404301 |
6 | 10500352101112121 |
7 | 451141265411266 |
oct | 56614562602221 |
9 | 12337703065084 |
10 | 3214343341201 |
11 | 102a217022473 |
12 | 43ab650b8641 |
13 | 1a415a622305 |
14 | b1809d0726d |
15 | 5892c01c051 |
hex | 2ec65cb0491 |
3214343341201 has 2 divisors, whose sum is σ = 3214343341202. Its totient is φ = 3214343341200.
The previous prime is 3214343341177. The next prime is 3214343341219. The reversal of 3214343341201 is 1021433434123.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 3159324947601 + 55018393600 = 1777449^2 + 234560^2 .
It is a cyclic number.
It is not a de Polignac number, because 3214343341201 - 215 = 3214343308433 is a prime.
It is not a weakly prime, because it can be changed into another prime (3214343341271) 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, 1607171670600 + 1607171670601.
It is an arithmetic number, because the mean of its divisors is an integer number (1607171670601).
Almost surely, 23214343341201 is an apocalyptic number.
It is an amenable number.
3214343341201 is a deficient number, since it is larger than the sum of its proper divisors (1).
3214343341201 is an equidigital number, since it uses as much as digits as its factorization.
3214343341201 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 20736, while the sum is 31.
Adding to 3214343341201 its reverse (1021433434123), we get a palindrome (4235776775324).
The spelling of 3214343341201 in words is "three trillion, two hundred fourteen billion, three hundred forty-three million, three hundred forty-one thousand, two hundred one".
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