Base | Representation |
---|---|
bin | 1011111011001111001111… |
… | …0001100101000111011111 |
3 | 1201102112012211022012120102 |
4 | 2332303303301211013133 |
5 | 3204313022443013332 |
6 | 43515415443211315 |
7 | 2522222522301533 |
oct | 276636361450737 |
9 | 51375184265512 |
10 | 13112330047967 |
11 | 41a59a9524387 |
12 | 1579310976b3b |
13 | 741644b1b217 |
14 | 3348d6859bc3 |
15 | 17b1360d9062 |
hex | becf3c651df |
13112330047967 has 2 divisors, whose sum is σ = 13112330047968. Its totient is φ = 13112330047966.
The previous prime is 13112330047907. The next prime is 13112330048003. The reversal of 13112330047967 is 76974003321131.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 13112330047967 - 210 = 13112330046943 is a prime.
It is a super-3 number, since 3×131123300479673 (a number of 40 digits) contains 333 as substring.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (13112330047907) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6556165023983 + 6556165023984.
It is an arithmetic number, because the mean of its divisors is an integer number (6556165023984).
Almost surely, 213112330047967 is an apocalyptic number.
13112330047967 is a deficient number, since it is larger than the sum of its proper divisors (1).
13112330047967 is an equidigital number, since it uses as much as digits as its factorization.
13112330047967 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 571536, while the sum is 47.
The spelling of 13112330047967 in words is "thirteen trillion, one hundred twelve billion, three hundred thirty million, forty-seven thousand, nine hundred sixty-seven".
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