Base | Representation |
---|---|
bin | 11101110101011000101101… |
… | …010111111001011100110011 |
3 | 122012120202010201122221021111 |
4 | 131311120231113321130303 |
5 | 114144234200101212042 |
6 | 1143021544412134151 |
7 | 36431516656641112 |
oct | 3565305527713463 |
9 | 565522121587244 |
10 | 131212012132147 |
11 | 3889877a181a61 |
12 | 12871925467357 |
13 | 582a31669189b |
14 | 24589a3b81c79 |
15 | 10281d3aee517 |
hex | 77562d5f9733 |
131212012132147 has 2 divisors, whose sum is σ = 131212012132148. Its totient is φ = 131212012132146.
The previous prime is 131212012132109. The next prime is 131212012132171. The reversal of 131212012132147 is 741231210212131.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 131212012132147 - 227 = 131211877914419 is a prime.
It is not a weakly prime, because it can be changed into another prime (131212012136147) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (29) of ones.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 65606006066073 + 65606006066074.
It is an arithmetic number, because the mean of its divisors is an integer number (65606006066074).
Almost surely, 2131212012132147 is an apocalyptic number.
131212012132147 is a deficient number, since it is larger than the sum of its proper divisors (1).
131212012132147 is an equidigital number, since it uses as much as digits as its factorization.
131212012132147 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 4032, while the sum is 31.
Adding to 131212012132147 its reverse (741231210212131), we get a palindrome (872443222344278).
The spelling of 131212012132147 in words is "one hundred thirty-one trillion, two hundred twelve billion, twelve million, one hundred thirty-two thousand, one hundred forty-seven".
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