Base | Representation |
---|---|
bin | 100101001100101010100… |
… | …0100100100000111101111 |
3 | 200002202002121200201100221 |
4 | 1022121111010210013233 |
5 | 1132230233203422002 |
6 | 14512342144232211 |
7 | 1035234662620564 |
oct | 112312504440757 |
9 | 20082077621327 |
10 | 5112438342127 |
11 | 16a119a188156 |
12 | 6a69a930b667 |
13 | 2b11407b2b5c |
14 | 13962c72a86b |
15 | 8cebd915c37 |
hex | 4a6551241ef |
5112438342127 has 2 divisors, whose sum is σ = 5112438342128. Its totient is φ = 5112438342126.
The previous prime is 5112438342101. The next prime is 5112438342133. The reversal of 5112438342127 is 7212438342115.
It is a happy number.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 5112438342127 - 215 = 5112438309359 is a prime.
It is a super-2 number, since 2×51124383421272 (a number of 26 digits) contains 22 as substring.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (5112438342527) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 2556219171063 + 2556219171064.
It is an arithmetic number, because the mean of its divisors is an integer number (2556219171064).
Almost surely, 25112438342127 is an apocalyptic number.
5112438342127 is a deficient number, since it is larger than the sum of its proper divisors (1).
5112438342127 is an equidigital number, since it uses as much as digits as its factorization.
5112438342127 is an evil number, because the sum of its binary digits is even.
The product of its digits is 322560, while the sum is 43.
The spelling of 5112438342127 in words is "five trillion, one hundred twelve billion, four hundred thirty-eight million, three hundred forty-two thousand, one hundred twenty-seven".
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