Given the relation $k=450nl+55n+205l+25$ where $k,n,l$ are all integers with $n,l<k$, if we know the upper limit for $k$,e.g. assume $k<=1000$ how many $n$ and $l$ could we find that satisfy the above relation? can the answer be given over a formula?
Asked
Active
Viewed 33 times
1 Answers
0
\begin{align} & k = 450 nl + 55n + 205 l + 25 \\ \implies & k - 55n -25 = (450n + 205)l \\ \implies & l = \dfrac{k - 55n -25}{450n + 205} \tag{1} \end{align}
Now you have $n, l < k$ and $k \leq 1000$, so for a given $k$ (say 500), all those integer values of $n$ which are less than 500 and gives an integer value of $l$ by satisfying (1) will be your solution. So depending on the value of $k$ you may have one / more solutions or no solutions at all (because of the integer constraint).
Nash J.
- 1,235
- 8
- 18