Base | Representation |
---|---|
bin | 11111000010101111000… |
… | …100101110011000000011 |
3 | 21112221021020100111010111 |
4 | 133002233010232120003 |
5 | 234422334200431042 |
6 | 4311555234412151 |
7 | 310056462400633 |
oct | 37025704563003 |
9 | 7487236314114 |
10 | 2133241030147 |
11 | 75277aa47927 |
12 | 2a552a131057 |
13 | 126218a42106 |
14 | 7536c5846c3 |
15 | 3a755646517 |
hex | 1f0af12e603 |
2133241030147 has 2 divisors, whose sum is σ = 2133241030148. Its totient is φ = 2133241030146.
The previous prime is 2133241030133. The next prime is 2133241030153. The reversal of 2133241030147 is 7410301423312.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 2133241030147 - 231 = 2131093546499 is a prime.
It is a super-3 number, since 3×21332410301473 (a number of 38 digits) contains 333 as substring.
It is not a weakly prime, because it can be changed into another prime (2133241030157) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 1066620515073 + 1066620515074.
It is an arithmetic number, because the mean of its divisors is an integer number (1066620515074).
Almost surely, 22133241030147 is an apocalyptic number.
2133241030147 is a deficient number, since it is larger than the sum of its proper divisors (1).
2133241030147 is an equidigital number, since it uses as much as digits as its factorization.
2133241030147 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 12096, while the sum is 31.
Adding to 2133241030147 its reverse (7410301423312), we get a palindrome (9543542453459).
The spelling of 2133241030147 in words is "two trillion, one hundred thirty-three billion, two hundred forty-one million, thirty thousand, one hundred forty-seven".
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