mingzhuqiuxia wrote:
Its a more complex problem when X is divide by two different int with two different remainder
for example if n is divided by 12 and remainder is 3 (we can write n = 12x + 3) and when n is divided by 5, its remainder is 1 (n = 5y + 1). Here x and y are quotients.
Lets equate both equations 12x+3 = 5y + 1,
[12x+2][/5]=y
now plug some integer value of x so that result is integer,
plugging x= 4 gives y = 50/5=10, plug 10 in n = 5(10) +1 = 51, this is one possible value of n true for both.