Base | Representation |
---|---|
bin | 110111010010000001100001… |
… | …011110010000001011011011 |
3 | 1011212212002222110102110002102 |
4 | 313102001201132100023123 |
5 | 223331430040331400012 |
6 | 2221032523340354015 |
7 | 102132434254345262 |
oct | 6722014136201333 |
9 | 1155762873373072 |
10 | 243131144012507 |
11 | 70518380419803 |
12 | 2332854024590b |
13 | a58822627ba2a |
14 | 44078783b95d9 |
15 | 1d195eb9e02c2 |
hex | dd20617902db |
243131144012507 has 2 divisors, whose sum is σ = 243131144012508. Its totient is φ = 243131144012506.
The previous prime is 243131144012501. The next prime is 243131144012527. The reversal of 243131144012507 is 705210441131342.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 243131144012507 - 26 = 243131144012443 is a prime.
It is a super-2 number, since 2×2431311440125072 (a number of 30 digits) contains 22 as substring.
It is not a weakly prime, because it can be changed into another prime (243131144012501) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 121565572006253 + 121565572006254.
It is an arithmetic number, because the mean of its divisors is an integer number (121565572006254).
Almost surely, 2243131144012507 is an apocalyptic number.
243131144012507 is a deficient number, since it is larger than the sum of its proper divisors (1).
243131144012507 is an equidigital number, since it uses as much as digits as its factorization.
243131144012507 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 80640, while the sum is 38.
Adding to 243131144012507 its reverse (705210441131342), we get a palindrome (948341585143849).
The spelling of 243131144012507 in words is "two hundred forty-three trillion, one hundred thirty-one billion, one hundred forty-four million, twelve thousand, five hundred seven".
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