Base | Representation |
---|---|
bin | 111001110110011100100001… |
… | …110101100001000111011011 |
3 | 1020100212021122120222112012211 |
4 | 321312130201311201013123 |
5 | 231322040404120401042 |
6 | 2301043331321415551 |
7 | 104406653640011635 |
oct | 7166344165410733 |
9 | 1210767576875184 |
10 | 254430135325147 |
11 | 7408423a4581a1 |
12 | 2465232163a5b7 |
13 | abc7874542252 |
14 | 46b86acb23a55 |
15 | 1e634a200d017 |
hex | e76721d611db |
254430135325147 has 2 divisors, whose sum is σ = 254430135325148. Its totient is φ = 254430135325146.
The previous prime is 254430135325103. The next prime is 254430135325153. The reversal of 254430135325147 is 741523531034452.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 254430135325147 - 231 = 254427987841499 is a prime.
It is a self number, because there is not a number n which added to its sum of digits gives 254430135325147.
It is not a weakly prime, because it can be changed into another prime (254430135305147) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 127215067662573 + 127215067662574.
It is an arithmetic number, because the mean of its divisors is an integer number (127215067662574).
Almost surely, 2254430135325147 is an apocalyptic number.
254430135325147 is a deficient number, since it is larger than the sum of its proper divisors (1).
254430135325147 is an equidigital number, since it uses as much as digits as its factorization.
254430135325147 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 6048000, while the sum is 49.
Adding to 254430135325147 its reverse (741523531034452), we get a palindrome (995953666359599).
The spelling of 254430135325147 in words is "two hundred fifty-four trillion, four hundred thirty billion, one hundred thirty-five million, three hundred twenty-five thousand, one hundred forty-seven".
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