Base | Representation |
---|---|
bin | 10011000101111010101… |
… | …010101011000011010001 |
3 | 11122102120010210202120021 |
4 | 103011322222223003101 |
5 | 132444010344011032 |
6 | 2442422434543441 |
7 | 163535103523015 |
oct | 23057252530321 |
9 | 4572503722507 |
10 | 1312023032017 |
11 | 466475885095 |
12 | 192341a37b81 |
13 | 96952c64479 |
14 | 47706280545 |
15 | 241de6e5097 |
hex | 1317aaab0d1 |
1312023032017 has 2 divisors, whose sum is σ = 1312023032018. Its totient is φ = 1312023032016.
The previous prime is 1312023031981. The next prime is 1312023032057. The reversal of 1312023032017 is 7102303202131.
It is a happy number.
It is a weak prime.
It can be written as a sum of positive squares in only one way, i.e., 1104010416961 + 208012615056 = 1050719^2 + 456084^2 .
It is a cyclic number.
It is not a de Polignac number, because 1312023032017 - 27 = 1312023031889 is a prime.
It is a junction number, because it is equal to n+sod(n) for n = 1312023031982 and 1312023032000.
It is not a weakly prime, because it can be changed into another prime (1312023032057) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 656011516008 + 656011516009.
It is an arithmetic number, because the mean of its divisors is an integer number (656011516009).
Almost surely, 21312023032017 is an apocalyptic number.
It is an amenable number.
1312023032017 is a deficient number, since it is larger than the sum of its proper divisors (1).
1312023032017 is an equidigital number, since it uses as much as digits as its factorization.
1312023032017 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 1512, while the sum is 25.
Adding to 1312023032017 its reverse (7102303202131), we get a palindrome (8414326234148).
The spelling of 1312023032017 in words is "one trillion, three hundred twelve billion, twenty-three million, thirty-two thousand, seventeen".
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