let n be a natural number such that the division n¡5 leaves a remainder 4 and the division n¡2 leaves a remainder 1 what must be the unit digit of n?
n=2k+1, so clearly n must be odd.
Now, it is also 5k+4, so k must be odd since otherwise 5k is even, and so is n.
Any odd multiple of 5 ends in 5, so n must end in 9.