Base | Representation |
---|---|
bin | 100101001100111011100… |
… | …1011001110101011000111 |
3 | 200002210120111100002111011 |
4 | 1022121313023032223013 |
5 | 1132232421142432011 |
6 | 14512515024023051 |
7 | 1035255111200155 |
oct | 112316713165307 |
9 | 20083514302434 |
10 | 5113010514631 |
11 | 16a1463159898 |
12 | 6a6b28a61a87 |
13 | 2b12021b7576 |
14 | 13968470c3d5 |
15 | 8d003c88621 |
hex | 4a6772ceac7 |
5113010514631 has 2 divisors, whose sum is σ = 5113010514632. Its totient is φ = 5113010514630.
The previous prime is 5113010514461. The next prime is 5113010514709. The reversal of 5113010514631 is 1364150103115.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 5113010514631 - 211 = 5113010512583 is a prime.
It is a super-2 number, since 2×51130105146312 (a number of 26 digits) contains 22 as substring.
It is a junction number, because it is equal to n+sod(n) for n = 5113010514593 and 5113010514602.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (5113010515631) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 2556505257315 + 2556505257316.
It is an arithmetic number, because the mean of its divisors is an integer number (2556505257316).
Almost surely, 25113010514631 is an apocalyptic number.
5113010514631 is a deficient number, since it is larger than the sum of its proper divisors (1).
5113010514631 is an equidigital number, since it uses as much as digits as its factorization.
5113010514631 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 5400, while the sum is 31.
Adding to 5113010514631 its reverse (1364150103115), we get a palindrome (6477160617746).
The spelling of 5113010514631 in words is "five trillion, one hundred thirteen billion, ten million, five hundred fourteen thousand, six hundred thirty-one".
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