"In this problem, \(a\) and \(b\) are integers.
If \(a\equiv 4\pmod{14}\) and \(a-b\equiv 10\pmod{14}\), then what is the remainder when \(a+b\) is divided by \(14\)?"
I found this (let \(x\) and \(y\) be positive intergers):
\(\begin{align*} a&=14x+14 \\ a-b&=14y+10 \\ 14x+4-b&=14y+10 \\ 14x-14y-6&=b \\ a+b&=14x+4+14x-14y-6 \\ a+b&=28x-14y-2 \\ \frac{a+b}{14}&=2x-y-\frac17 \\ \end{align*}\)
I'm not quite sure how to get the remainder since \(-\frac17\) is negative.
Adding the two congruences, we get 2a≡14(mod14), so a≡7(mod14). Substituting this into either of the original congruences, we get 7−b≡10(mod14), so b≡−3(mod14). Therefore, a+b ≡7−3 ≡ 4 (mod14). The remainder when a + b is divided by 14 is 4.
+397 The remainder should be an integer and you have that one line from the bottom.
a + b = 28x - 14y - 2 = 14k - 2, (where k = 28x - 14y).
The remainder after division by 14 is -2.
If you don't like the negative sign you can simply add 14 to get a remainder of 12.
Check it out with some random substitutions for a and b if you wish.
For example, let a = 74 (= 5*14 + 4),
let a - b = 38 (= 2*14 +10),
then b = 74 - 38 = 36,
so a + b = 74 + 36 = 110 = 98 + 12 = 7*14 + 12, remainder 12,
(or 110 = 112 - 2 = 8*14 - 2, remainder -2).
