Base | Representation |
---|---|
bin | 10000011110010110011… |
… | …110111110001110010001 |
3 | 11000020011001221122001102 |
4 | 100132112132332032101 |
5 | 122022020340203213 |
6 | 2224025143314145 |
7 | 144535343402432 |
oct | 20362636761621 |
9 | 4006131848042 |
10 | 1132101100433 |
11 | 3a7137614966 |
12 | 1634aa44b955 |
13 | 829ab555064 |
14 | 3cb189a7489 |
15 | 1e6adcad158 |
hex | 107967be391 |
1132101100433 has 2 divisors, whose sum is σ = 1132101100434. Its totient is φ = 1132101100432.
The previous prime is 1132101100421. The next prime is 1132101100453. The reversal of 1132101100433 is 3340011012311.
It is an a-pointer prime, because the next prime (1132101100453) can be obtained adding 1132101100433 to its sum of digits (20).
It is a weak prime.
It can be written as a sum of positive squares in only one way, i.e., 924795725569 + 207305374864 = 961663^2 + 455308^2 .
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-1132101100433 is a prime.
It is not a weakly prime, because it can be changed into another prime (1132101100453) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 566050550216 + 566050550217.
It is an arithmetic number, because the mean of its divisors is an integer number (566050550217).
Almost surely, 21132101100433 is an apocalyptic number.
It is an amenable number.
1132101100433 is a deficient number, since it is larger than the sum of its proper divisors (1).
1132101100433 is an equidigital number, since it uses as much as digits as its factorization.
1132101100433 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 216, while the sum is 20.
Adding to 1132101100433 its reverse (3340011012311), we get a palindrome (4472112112744).
The spelling of 1132101100433 in words is "one trillion, one hundred thirty-two billion, one hundred one million, one hundred thousand, four hundred thirty-three".
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