Base | Representation |
---|---|
bin | 11110001111100100011100… |
… | …011111111101011100111111 |
3 | 122102221202110121110220200011 |
4 | 132033210130133331130333 |
5 | 114413224141001030142 |
6 | 1150520320151244051 |
7 | 40005515341163335 |
oct | 3617443437753477 |
9 | 572852417426604 |
10 | 133011320330047 |
11 | 39421869801475 |
12 | 12b0259a079627 |
13 | 592abb54a7c23 |
14 | 24bbad470bb55 |
15 | 1059de2a54417 |
hex | 78f91c7fd73f |
133011320330047 has 2 divisors, whose sum is σ = 133011320330048. Its totient is φ = 133011320330046.
The previous prime is 133011320330023. The next prime is 133011320330051. The reversal of 133011320330047 is 740033023110331.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 133011320330047 - 27 = 133011320329919 is a prime.
It is a super-4 number, since 4×1330113203300474 (a number of 58 digits) contains 4444 as substring. Note that it is a super-d number also for d = 2.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (133011320330947) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 66505660165023 + 66505660165024.
It is an arithmetic number, because the mean of its divisors is an integer number (66505660165024).
Almost surely, 2133011320330047 is an apocalyptic number.
133011320330047 is a deficient number, since it is larger than the sum of its proper divisors (1).
133011320330047 is an equidigital number, since it uses as much as digits as its factorization.
133011320330047 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 13608, while the sum is 31.
Adding to 133011320330047 its reverse (740033023110331), we get a palindrome (873044343440378).
The spelling of 133011320330047 in words is "one hundred thirty-three trillion, eleven billion, three hundred twenty million, three hundred thirty thousand, forty-seven".
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