Base | Representation |
---|---|
bin | 111100000110110000011… |
… | …010100001001011111111 |
3 | 112121212100022011111000211 |
4 | 330012300122201023333 |
5 | 1020133101332010101 |
6 | 12441253522133251 |
7 | 604261511610442 |
oct | 74066032411377 |
9 | 15555308144024 |
10 | 4130423313151 |
11 | 135278030a183 |
12 | 5686064b1227 |
13 | 23c6605b5005 |
14 | 103cb0b1b659 |
15 | 72696089c51 |
hex | 3c1b06a12ff |
4130423313151 has 2 divisors, whose sum is σ = 4130423313152. Its totient is φ = 4130423313150.
The previous prime is 4130423313119. The next prime is 4130423313187. The reversal of 4130423313151 is 1513133240314.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 4130423313151 - 25 = 4130423313119 is a prime.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (4130423311151) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 2065211656575 + 2065211656576.
It is an arithmetic number, because the mean of its divisors is an integer number (2065211656576).
Almost surely, 24130423313151 is an apocalyptic number.
4130423313151 is a deficient number, since it is larger than the sum of its proper divisors (1).
4130423313151 is an equidigital number, since it uses as much as digits as its factorization.
4130423313151 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 12960, while the sum is 31.
Adding to 4130423313151 its reverse (1513133240314), we get a palindrome (5643556553465).
The spelling of 4130423313151 in words is "four trillion, one hundred thirty billion, four hundred twenty-three million, three hundred thirteen thousand, one hundred fifty-one".
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