have the following equation: $(7i_2+13j_2+15)+(i_1+j_1+2)$ must be divisible by 18. The variables $ i_1, i_2$ and $j_1, j_2$ can take values 0,1,2,3,... if we pair the variables $(i_1,j_1)$ and $(i_2,j_2)$ then in a given interval, say 0-1000 (i.e. all variables accept only values in this interval), How many such solution pairs can there be?
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For each of the $1001^3$ values of $i_2, j_2, i_1$, there are about $\frac {1001}{18}$ solutions for $j_1$. Thus the answer will be near $\frac {1001^4}{18}$. However, vagaries at the boundaries can make it tricky to get the exact count. If you had chosen an interval (such as 0 to 1007) whose size $N$ is divisible by $18$, then the number of solutions would be exactly $$\frac {N^4}{18}$$. – Paul Sinclair Sep 08 '17 at 00:42
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@PaulSinclair I wrote a small java program that generates such $k$ from 0 up to 180. after removing duplicates (there are evidently duplicate values of $k$) counting the generated $k$ are only 169! I think your calculation does not consider the constrain of congruence completely. – bijan karimi Sep 08 '17 at 22:49
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@PaulSinclair sorry Paul, I had forgotten to mention a side condition, that mislead you. takinh the side condition into account, then the number reduces drastically. but your analysis is basically correct. – bijan karimi Sep 09 '17 at 01:34
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Well, your comment about "pairing the variables" and counting "solution pairs" didn't make much sense, as a solution to the problem as stated is a set of values for all four variables, which is what I was counting. So I assumed there was something else, but since you didn't give the rest, I couldn't account for it. By the way, your test using 0 to 180 includes 181 values, so it isn't quite divisible by 18. – Paul Sinclair Sep 09 '17 at 02:20