Base | Representation |
---|---|
bin | 1011110101000001011100… |
… | …1101010101011111000011 |
3 | 1201001022121011110020001122 |
4 | 2331100113031111133003 |
5 | 3201040331342003011 |
6 | 43354400051422455 |
7 | 2511422442632612 |
oct | 275202715253703 |
9 | 51038534406048 |
10 | 13005550344131 |
11 | 416469412a464 |
12 | 15606905b4a2b |
13 | 734558848007 |
14 | 32d687225879 |
15 | 178486a9d3db |
hex | bd4173557c3 |
13005550344131 has 2 divisors, whose sum is σ = 13005550344132. Its totient is φ = 13005550344130.
The previous prime is 13005550344019. The next prime is 13005550344173. The reversal of 13005550344131 is 13144305550031.
It is a strong prime.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-13005550344131 is a prime.
It is a super-2 number, since 2×130055503441312 (a number of 27 digits) contains 22 as substring.
It is a Sophie Germain prime.
It is a junction number, because it is equal to n+sod(n) for n = 13005550344091 and 13005550344100.
It is not a weakly prime, because it can be changed into another prime (13005550341131) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6502775172065 + 6502775172066.
It is an arithmetic number, because the mean of its divisors is an integer number (6502775172066).
Almost surely, 213005550344131 is an apocalyptic number.
13005550344131 is a deficient number, since it is larger than the sum of its proper divisors (1).
13005550344131 is an equidigital number, since it uses as much as digits as its factorization.
13005550344131 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 54000, while the sum is 35.
Adding to 13005550344131 its reverse (13144305550031), we get a palindrome (26149855894162).
The spelling of 13005550344131 in words is "thirteen trillion, five billion, five hundred fifty million, three hundred forty-four thousand, one hundred thirty-one".
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