Base | Representation |
---|---|
bin | 111110000010010111… |
… | …0100011100001001011 |
3 | 110201212120010210210201 |
4 | 1330010232203201023 |
5 | 4140320131033042 |
6 | 141111341353031 |
7 | 12424253233651 |
oct | 1740456434113 |
9 | 421776123721 |
10 | 133223299147 |
11 | 51555096524 |
12 | 219a0268777 |
13 | c741910333 |
14 | 663b5dadd1 |
15 | 36ead151b7 |
hex | 1f04ba384b |
133223299147 has 2 divisors, whose sum is σ = 133223299148. Its totient is φ = 133223299146.
The previous prime is 133223299117. The next prime is 133223299151. The reversal of 133223299147 is 741992322331.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 133223299147 - 211 = 133223297099 is a prime.
It is a super-2 number, since 2×1332232991472 (a number of 23 digits) contains 22 as substring.
It is a junction number, because it is equal to n+sod(n) for n = 133223299097 and 133223299106.
It is not a weakly prime, because it can be changed into another prime (133223299117) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 66611649573 + 66611649574.
It is an arithmetic number, because the mean of its divisors is an integer number (66611649574).
Almost surely, 2133223299147 is an apocalyptic number.
133223299147 is a deficient number, since it is larger than the sum of its proper divisors (1).
133223299147 is an equidigital number, since it uses as much as digits as its factorization.
133223299147 is an evil number, because the sum of its binary digits is even.
The product of its digits is 489888, while the sum is 46.
The spelling of 133223299147 in words is "one hundred thirty-three billion, two hundred twenty-three million, two hundred ninety-nine thousand, one hundred forty-seven".
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