Base | Representation |
---|---|
bin | 1100111101000100100000… |
… | …0110001101010001111111 |
3 | 1212102122111112120120211121 |
4 | 3033101020012031101333 |
5 | 3331330310202442401 |
6 | 50143142535331411 |
7 | 3000021530046232 |
oct | 317211006152177 |
9 | 55378445516747 |
10 | 14243321140351 |
11 | 45a1616508816 |
12 | 172054b520b67 |
13 | 7c41a73ca184 |
14 | 373547c3b419 |
15 | 19a77c6d5ca1 |
hex | cf44818d47f |
14243321140351 has 2 divisors, whose sum is σ = 14243321140352. Its totient is φ = 14243321140350.
The previous prime is 14243321140337. The next prime is 14243321140427. The reversal of 14243321140351 is 15304112334241.
14243321140351 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.
It is a weak prime.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-14243321140351 is a prime.
It is a super-2 number, since 2×142433211403512 (a number of 27 digits) contains 22 as substring.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (14243321147351) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 7121660570175 + 7121660570176.
It is an arithmetic number, because the mean of its divisors is an integer number (7121660570176).
Almost surely, 214243321140351 is an apocalyptic number.
14243321140351 is a deficient number, since it is larger than the sum of its proper divisors (1).
14243321140351 is an equidigital number, since it uses as much as digits as its factorization.
14243321140351 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 34560, while the sum is 34.
Adding to 14243321140351 its reverse (15304112334241), we get a palindrome (29547433474592).
The spelling of 14243321140351 in words is "fourteen trillion, two hundred forty-three billion, three hundred twenty-one million, one hundred forty thousand, three hundred fifty-one".
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