Base | Representation |
---|---|
bin | 10000100111000010100… |
… | …110100101111010110011 |
3 | 11001010020022020022212121 |
4 | 100213002212211322303 |
5 | 122200122343331011 |
6 | 2232211040335111 |
7 | 145315505241643 |
oct | 20470246457263 |
9 | 4033208208777 |
10 | 1141431230131 |
11 | 400095199455 |
12 | 165273014497 |
13 | 8383752216a |
14 | 3d361b96523 |
15 | 1ea57e4ae71 |
hex | 109c29a5eb3 |
1141431230131 has 2 divisors, whose sum is σ = 1141431230132. Its totient is φ = 1141431230130.
The previous prime is 1141431230063. The next prime is 1141431230207. The reversal of 1141431230131 is 1310321341411.
It is a weak prime.
It is an emirp because it is prime and its reverse (1310321341411) is a distict prime.
It is a cyclic number.
It is not a de Polignac number, because 1141431230131 - 27 = 1141431230003 is a prime.
It is a super-2 number, since 2×11414312301312 (a number of 25 digits) contains 22 as substring.
It is a junction number, because it is equal to n+sod(n) for n = 1141431230096 and 1141431230105.
It is not a weakly prime, because it can be changed into another prime (1141431530131) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 570715615065 + 570715615066.
It is an arithmetic number, because the mean of its divisors is an integer number (570715615066).
Almost surely, 21141431230131 is an apocalyptic number.
1141431230131 is a deficient number, since it is larger than the sum of its proper divisors (1).
1141431230131 is an equidigital number, since it uses as much as digits as its factorization.
1141431230131 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 864, while the sum is 25.
Adding to 1141431230131 its reverse (1310321341411), we get a palindrome (2451752571542).
The spelling of 1141431230131 in words is "one trillion, one hundred forty-one billion, four hundred thirty-one million, two hundred thirty thousand, one hundred thirty-one".
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