Base | Representation |
---|---|
bin | 1100111011010101000101… |
… | …0100000100111110111011 |
3 | 1212022210022001101001001221 |
4 | 3032311101110010332323 |
5 | 3330333031204002001 |
6 | 50121314553252511 |
7 | 2664612401050501 |
oct | 316652124047673 |
9 | 55283261331057 |
10 | 14213411000251 |
11 | 458a967a6a871 |
12 | 17167a2704137 |
13 | 7c141b838318 |
14 | 371d0b704671 |
15 | 199acb90c1a1 |
hex | ced51504fbb |
14213411000251 has 2 divisors, whose sum is σ = 14213411000252. Its totient is φ = 14213411000250.
The previous prime is 14213411000227. The next prime is 14213411000263. The reversal of 14213411000251 is 15200011431241.
It is a happy number.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 14213411000251 - 25 = 14213411000219 is a prime.
It is a super-2 number, since 2×142134110002512 (a number of 27 digits) contains 22 as substring.
It is not a weakly prime, because it can be changed into another prime (14213411000351) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 7106705500125 + 7106705500126.
It is an arithmetic number, because the mean of its divisors is an integer number (7106705500126).
Almost surely, 214213411000251 is an apocalyptic number.
14213411000251 is a deficient number, since it is larger than the sum of its proper divisors (1).
14213411000251 is an equidigital number, since it uses as much as digits as its factorization.
14213411000251 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 960, while the sum is 25.
Adding to 14213411000251 its reverse (15200011431241), we get a palindrome (29413422431492).
The spelling of 14213411000251 in words is "fourteen trillion, two hundred thirteen billion, four hundred eleven million, two hundred fifty-one", and thus it is an aban number.
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