I was interested in listing all the possible integer solutions to
f(n/10)-f(n/11) = 1 (eq1)
Where f(x)=floor(x) is the floor function, relating each real number x to the greatest integer z less or equal to x.
The floor function wasn't so easy to deal with, as it seems at first sight. I replaced n/10 = x.. and took 10x/11, having a similar equation:
f(x)-f(10x/11) = 1
The solution set was then easily verified, X = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}..
But didn't mean the possible solutions to (eq1) would resume to:
Y= 10*X = {10, 20, 30, 40, 50, 60, 70, 80, 90, 100, 110}.
I computed every solution to the equation (shown in the figure) with the support of an algorithm.
I realized then, the relation with another equation by taking a look thru the floor function identities...
–11x + 10y = 110 – n ,
where x = n mod 10, and y = n mod 11.
Seems like a diophantine approximation involved. Of course there are theorems to help with solutions of diophantine equations... But...
What if we have an equation:
f(x/a) + f(x/b) = c, where x is the variable and a, b, c are positive integers, where f(x)=floor(x) is once again the floor function.
How can we compute the possible solutions for x integer?
Is there any property that we can operate with floor functions? I suppose not cuz the function isn't continuous.
This subject just caught my attention. Maybe it's easier than seems. Any clue/tips?
Peter, your approach is excellent.
It seems to me that an imperfection is there:
In the case k=1, the missing values are gradually 1 piece, 2 pieces, 3 pieces, ...
i.e. r ≠ 10,20,21,30,31,32,...,90,...,98,100,...,109.
These excluded values of r give [r/10]-[r/11]=1, thus they are convenient for the case k=0.
Peter, as I wrote already, I find your approach excellent. In mathematics, it is common to make a mistake, and it is not a shame. Only those who do nothing do not make mistakes. You wrote your solution too quickly in your half-inch high text window, it is all :o)
The first time already, I followed your solution just till the part that you do not like [do we have the same taste? :) ], then continued as follows:
All numbers r such that 10m ≤ r < 11m for m=1,...,10 give [r/10]-[r/11]=1. These numbers form a set S1.
All other numbers r from the set {1,2,...,109} give [r/10]-[r/11]=0, denote this set S0.
Then I compared our results.
To conclude:
The numbers r∈S1 together with k=0 give solutions n∈S1.
The numbers r∈S0 with k=1 give solutions n∈{110+r; r∈S0}.
Thank you both for the contributions.
I am still intrigued with operations with those floor and ceiling functions.. I mean, how to compute the amount of possible solutions to x as positive integer, with something like [x/a] + [x/b] = c, with a, b, c given positive integers.
Thank you once again.
Peter, your last suggestion is connected with a,b coprimes - or not necessarily?
Dear Kelvin,
The equation [x/a] + [y/b] = c, where [ ] is the floor function, has the
general solution: x = ma + r and y = (c-m)b + s,
where m is any arbitrary integer, 0 ≤ r
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