Go through my method in this post.
Edit: Unless you meant this as a challenge problem of course, in which case this thread is under the wrong sub-forum.
Hi!
I have chosen a positive integer. On division by 14 my integer gives remainder 6. On division by 15 my integer gives remainder 2. On division by 13 my integer gives remainder 12. What is the smallest possible positive integer I could have chosen?
Good luck!
x = 6(mod14)
x=14a +6
x=2(mod15)
14a+6 = 2(mod15)
14a=-4(mod15)
14a=11(mod15)
14a=15b+11
combining my equations.... x=15b+17
15b+17=12(mod13)
15b=-5(mod13)
3b=-1(mod13)
3b=13c-1
x=15b+17
x=5(13c-1)+17
x=65c+12
.....from here i am stuck, i seem to be going roundin circles?
oh ok, thank you, basically i was confused as to why it was only at this point that we put c=0, rather than say take b=0 in x= 210b+62.... Then i realised that because at that point, not all parts of the question had been considered, i.e. x=12(mod13) hadnt been processed.... this makes sense now :-)