Base | Representation |
---|---|
bin | 10011000101111010001… |
… | …001110111101111011001 |
3 | 11122102112112121220021002 |
4 | 103011322021313233121 |
5 | 132444001143330431 |
6 | 2442421530354345 |
7 | 163534640450501 |
oct | 23057211675731 |
9 | 4572475556232 |
10 | 1312014433241 |
11 | 466470a41766 |
12 | 19233ab8b9b5 |
13 | 96951243635 |
14 | 47705082a01 |
15 | 241dda973cb |
hex | 1317a277bd9 |
1312014433241 has 2 divisors, whose sum is σ = 1312014433242. Its totient is φ = 1312014433240.
The previous prime is 1312014433157. The next prime is 1312014433277. The reversal of 1312014433241 is 1423344102131.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 1066667577616 + 245346855625 = 1032796^2 + 495325^2 .
It is a cyclic number.
It is not a de Polignac number, because 1312014433241 - 222 = 1312010238937 is a prime.
It is a Sophie Germain prime.
It is a Curzon number.
It is not a weakly prime, because it can be changed into another prime (1312014433291) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 656007216620 + 656007216621.
It is an arithmetic number, because the mean of its divisors is an integer number (656007216621).
Almost surely, 21312014433241 is an apocalyptic number.
It is an amenable number.
1312014433241 is a deficient number, since it is larger than the sum of its proper divisors (1).
1312014433241 is an equidigital number, since it uses as much as digits as its factorization.
1312014433241 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 6912, while the sum is 29.
Adding to 1312014433241 its reverse (1423344102131), we get a palindrome (2735358535372).
The spelling of 1312014433241 in words is "one trillion, three hundred twelve billion, fourteen million, four hundred thirty-three thousand, two hundred forty-one".
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