Base | Representation |
---|---|
bin | 101111101011110100… |
… | …101100100100000001 |
3 | 11220011021220101010021 |
4 | 233223310230210001 |
5 | 1314324440200113 |
6 | 35304345133441 |
7 | 3461545112056 |
oct | 575364544401 |
9 | 156137811107 |
10 | 51201100033 |
11 | 1a794760306 |
12 | 9b0b18a281 |
13 | 4a9c85aac5 |
14 | 269a06902d |
15 | 14ea04b48d |
hex | bebd2c901 |
51201100033 has 2 divisors, whose sum is σ = 51201100034. Its totient is φ = 51201100032.
The previous prime is 51201100019. The next prime is 51201100037. The reversal of 51201100033 is 33000110215.
51201100033 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 42844032144 + 8357067889 = 206988^2 + 91417^2 .
It is a cyclic number.
It is not a de Polignac number, because 51201100033 - 25 = 51201100001 is a prime.
It is not a weakly prime, because it can be changed into another prime (51201100037) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 25600550016 + 25600550017.
It is an arithmetic number, because the mean of its divisors is an integer number (25600550017).
Almost surely, 251201100033 is an apocalyptic number.
It is an amenable number.
51201100033 is a deficient number, since it is larger than the sum of its proper divisors (1).
51201100033 is an equidigital number, since it uses as much as digits as its factorization.
51201100033 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 90, while the sum is 16.
Adding to 51201100033 its reverse (33000110215), we get a palindrome (84201210248).
The spelling of 51201100033 in words is "fifty-one billion, two hundred one million, one hundred thousand, thirty-three".
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