Base | Representation |
---|---|
bin | 1111100001101100111010… |
… | …10000010011111100111011 |
3 | 11110220001220121100021221111 |
4 | 13300312131100103330323 |
5 | 13433401021313203333 |
6 | 200341121504021151 |
7 | 10122526425552115 |
oct | 760663520237473 |
9 | 143801817307844 |
10 | 34143333334843 |
11 | a974119609647 |
12 | 39b52593017b7 |
13 | 1608919580b45 |
14 | 8607908d03b5 |
15 | 3e322e4404cd |
hex | 1f0d9d413f3b |
34143333334843 has 2 divisors, whose sum is σ = 34143333334844. Its totient is φ = 34143333334842.
The previous prime is 34143333334807. The next prime is 34143333334867. The reversal of 34143333334843 is 34843333334143.
It is a strong prime.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-34143333334843 is a prime.
It is a super-2 number, since 2×341433333348432 (a number of 28 digits) contains 22 as substring.
It is not a weakly prime, because it can be changed into another prime (34143333331843) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 17071666667421 + 17071666667422.
It is an arithmetic number, because the mean of its divisors is an integer number (17071666667422).
Almost surely, 234143333334843 is an apocalyptic number.
34143333334843 is a deficient number, since it is larger than the sum of its proper divisors (1).
34143333334843 is an equidigital number, since it uses as much as digits as its factorization.
34143333334843 is an evil number, because the sum of its binary digits is even.
The product of its digits is 13436928, while the sum is 49.
Adding to 34143333334843 its reverse (34843333334143), we get a palindrome (68986666668986).
The spelling of 34143333334843 in words is "thirty-four trillion, one hundred forty-three billion, three hundred thirty-three million, three hundred thirty-four thousand, eight hundred forty-three".
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