Compute the smallest positive integer k such that the fraction
(7k+100)/(5k-3)
is reducible.
The fraction being reducible is equivalent to the numerator and denominator having a common factor. That common factor must divide any integer linear combination. In particular the common factor must divide 5*(7k+100)-7*(5k-3)=521.
521 is prime, therefore each of 7k+100 and 5k-3 must be divisible by 521. I will focus on 5k-3. Then I can say 5k-3 = 521m for some m. Take this mod 5; then 2 = m mod 5. Obviously the smallest m is m=2.
Then 5k-3=521*2 yields k=209. Check the fraction: (7k+100)/(5k-3) = (7*209+100)/(5*209-3) = 1563/1042; this fraction is indeed reducible: 1563/1042 = (3*521)/(2*521) = 3/2.
Solution
