Base | Representation |
---|---|
bin | 110000000001111110… |
… | …0111010100110101001 |
3 | 100212020101111002020022 |
4 | 1200003330322212221 |
5 | 3142220222443213 |
6 | 115215010210225 |
7 | 10311020262302 |
oct | 1400374724651 |
9 | 325211432208 |
10 | 103145515433 |
11 | 3a81aa55434 |
12 | 17ba7283975 |
13 | 995a3b288c |
14 | 4dc6b057a9 |
15 | 2a3a474308 |
hex | 1803f3a9a9 |
103145515433 has 2 divisors, whose sum is σ = 103145515434. Its totient is φ = 103145515432.
The previous prime is 103145515409. The next prime is 103145515463. The reversal of 103145515433 is 334515541301.
It is a weak prime.
It can be written as a sum of positive squares in only one way, i.e., 90858236329 + 12287279104 = 301427^2 + 110848^2 .
It is a cyclic number.
It is not a de Polignac number, because 103145515433 - 220 = 103144466857 is a prime.
It is a junction number, because it is equal to n+sod(n) for n = 103145515393 and 103145515402.
It is not a weakly prime, because it can be changed into another prime (103145515463) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 51572757716 + 51572757717.
It is an arithmetic number, because the mean of its divisors is an integer number (51572757717).
Almost surely, 2103145515433 is an apocalyptic number.
It is an amenable number.
103145515433 is a deficient number, since it is larger than the sum of its proper divisors (1).
103145515433 is an equidigital number, since it uses as much as digits as its factorization.
103145515433 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 54000, while the sum is 35.
The spelling of 103145515433 in words is "one hundred three billion, one hundred forty-five million, five hundred fifteen thousand, four hundred thirty-three".
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