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henpen
How many integer n exist so that \[ \frac{n}{100-n} \] is also an integer?
What is the general strategy to be adopted for 'How many integer n exist so that f(n) is also an integer?', other than bursts of insight?
i think, trial error is the best way for this case
suppose that for some integer k, \[k=\frac{ n }{ 100-n }\] rearranging the equation we get \[n=\frac{ 100k }{ k+1 }\] and since k and k+1 are relatively prime, we conclude that \[(k+1)|100\]
What does the notation (k+1)|100 mean?
when k = 0, n = 0. when k = 1, n = 50 when k = 3, n = 75
there are more values for k, but the solution set is finite.