Base | Representation |
---|---|
bin | 100101001100001000111… |
… | …1000101000110011111111 |
3 | 200002122012001200000200101 |
4 | 1022120101320220303333 |
5 | 1132220431123022033 |
6 | 14512034240341531 |
7 | 1035165024600343 |
oct | 112302170506377 |
9 | 20078161600611 |
10 | 5111311142143 |
11 | 16a0770985118 |
12 | 6a67339148a7 |
13 | 2b0cc10ca80b |
14 | 139562b4b623 |
15 | 8ce549aa97d |
hex | 4a611e28cff |
5111311142143 has 2 divisors, whose sum is σ = 5111311142144. Its totient is φ = 5111311142142.
The previous prime is 5111311142141. The next prime is 5111311142149. The reversal of 5111311142143 is 3412411131115.
It is a happy number.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 5111311142143 - 21 = 5111311142141 is a prime.
It is a super-2 number, since 2×51113111421432 (a number of 26 digits) contains 22 as substring.
Together with 5111311142141, it forms a pair of twin primes.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (5111311142141) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 2555655571071 + 2555655571072.
It is an arithmetic number, because the mean of its divisors is an integer number (2555655571072).
Almost surely, 25111311142143 is an apocalyptic number.
5111311142143 is a deficient number, since it is larger than the sum of its proper divisors (1).
5111311142143 is an equidigital number, since it uses as much as digits as its factorization.
5111311142143 is an evil number, because the sum of its binary digits is even.
The product of its digits is 1440, while the sum is 28.
Adding to 5111311142143 its reverse (3412411131115), we get a palindrome (8523722273258).
The spelling of 5111311142143 in words is "five trillion, one hundred eleven billion, three hundred eleven million, one hundred forty-two thousand, one hundred forty-three".
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