Base | Representation |
---|---|
bin | 1100111011101110001000… |
… | …0011100011001101110011 |
3 | 1212100102122121111122100012 |
4 | 3032323202003203031303 |
5 | 3330440313342100001 |
6 | 50124350045445135 |
7 | 2665241120321552 |
oct | 316734203431563 |
9 | 55312577448305 |
10 | 14220134331251 |
11 | 45927a8125972 |
12 | 1717b5a2707ab |
13 | 7c1c50721aa4 |
14 | 372388601d99 |
15 | 199d71cc52bb |
hex | ceee20e3373 |
14220134331251 has 2 divisors, whose sum is σ = 14220134331252. Its totient is φ = 14220134331250.
The previous prime is 14220134331239. The next prime is 14220134331257. The reversal of 14220134331251 is 15213343102241.
It is a happy number.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 14220134331251 - 26 = 14220134331187 is a prime.
It is a Sophie Germain prime.
It is not a weakly prime, because it can be changed into another prime (14220134331257) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 7110067165625 + 7110067165626.
It is an arithmetic number, because the mean of its divisors is an integer number (7110067165626).
Almost surely, 214220134331251 is an apocalyptic number.
14220134331251 is a deficient number, since it is larger than the sum of its proper divisors (1).
14220134331251 is an equidigital number, since it uses as much as digits as its factorization.
14220134331251 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 17280, while the sum is 32.
Adding to 14220134331251 its reverse (15213343102241), we get a palindrome (29433477433492).
The spelling of 14220134331251 in words is "fourteen trillion, two hundred twenty billion, one hundred thirty-four million, three hundred thirty-one thousand, two hundred fifty-one".
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