Let's say the two remainders are r and R, so we have that

z= 12*x + r

z= 24*y + R

where 0<r<11 and 0<R<23.

Subtracting one equation from the another we obtain

0= 12*(x-2*y) + r - R => R= 12*(x-2*y) + r

Now it must be 0<=(x-2*y)<2 in order to be 0<R<23 = > R=r or R=12+r

In our case r=9 so It could be R=9 or R=21 .. but only one of these two choices.

@whiplash2411

You are right, remainders are good guys: if you want to find the remainder of a sum you can calculate the sum of the remainders ( ... and eventually calculate again the remainder if necessary ) BUT you cannot mix the numbers you are dividing by.

**Quote:**

Similarly here, we have \(\frac{12x}{24}\) which can be reduced to \(\frac{x}{2}\) . x can either be an odd number or even number and subsequently leave a reminder of 1 or 0.

Then we have \(\frac{9}{24}\) which will leave a reminder of 9.

You cannot consider the sum of the remainder of \(\frac{9}{24}\) and the remainder of \(\frac{x}{2}\) ... it will be the remainder of z divided by what... 24 or 2 ?

So following your approach if x is even then the remainder of \(\frac{12x}{24}\) will be 0, if x is odd the remainder will be

12 ( not 1).

_________________

\(Hugh\)