Base | Representation |
---|---|
bin | 1011000010001000010100… |
… | …0101110000110001110011 |
3 | 1120221201210121122101210022 |
4 | 2300202011011300301303 |
5 | 3042224214423233042 |
6 | 41445001125154055 |
7 | 2361310645036214 |
oct | 260420505606163 |
9 | 46851717571708 |
10 | 12131220524147 |
11 | 39579047a887a |
12 | 143b14125b92b |
13 | 69cc7a3c374a |
14 | 2dd2231db40b |
15 | 160862ec5cd2 |
hex | b0885170c73 |
12131220524147 has 2 divisors, whose sum is σ = 12131220524148. Its totient is φ = 12131220524146.
The previous prime is 12131220524069. The next prime is 12131220524167. The reversal of 12131220524147 is 74142502213121.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 12131220524147 - 224 = 12131203746931 is a prime.
It is a super-2 number, since 2×121312205241472 (a number of 27 digits) contains 22 as substring.
It is not a weakly prime, because it can be changed into another prime (12131220524167) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6065610262073 + 6065610262074.
It is an arithmetic number, because the mean of its divisors is an integer number (6065610262074).
Almost surely, 212131220524147 is an apocalyptic number.
12131220524147 is a deficient number, since it is larger than the sum of its proper divisors (1).
12131220524147 is an equidigital number, since it uses as much as digits as its factorization.
12131220524147 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 26880, while the sum is 35.
Adding to 12131220524147 its reverse (74142502213121), we get a palindrome (86273722737268).
The spelling of 12131220524147 in words is "twelve trillion, one hundred thirty-one billion, two hundred twenty million, five hundred twenty-four thousand, one hundred forty-seven".
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