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"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.

 Jun 13, 2023
 #1
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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.

 Jun 13, 2023
 #2
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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).

 Jun 13, 2023

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