Remainders of sums and differences

Start from the division algorithm. Every whole number $n$ divided by a divisor $d>0$ can be written uniquely as

where: $q$ is the quotient and $r$ is the remainder. This is the only tool this section uses — no new machinery, just the same idea applied to two numbers at once and then combined.

Adding two numbers with the same divisor. If $a = dq_1+r_1$ and $b = dq_2+r_2$ (same $d$), then

where: $r_1, r_2$ are the individual remainders, each already satisfying $0 \le r_1,r_2 < d$. The catch is that $r_1+r_2$ can itself be $d$ or larger — up to $2d-2$ in the worst case — so it is not automatically a valid remainder. Whenever $r_1+r_2 \ge d$, subtract $d$ once (and add $1$ to the quotient part) to land back in range; since $r_1+r_2<2d$, one subtraction is always enough.

Subtracting two numbers with the same divisor. Similarly $a-b = d(q_1-q_2)+(r_1-r_2)$, and if $r_1-r_2$ turns out negative, add $d$ once (and subtract $1$ from the quotient part) to bring it back into $0$ to $d-1$. This shortcut is exactly why the rule is worth learning: for two large numbers, working out their individual remainders once and then just adding or subtracting those two small numbers is far quicker than dividing the full sum or difference from scratch, and it is the standard move behind almost every "find the remainder" puzzle you will meet in this chapter. The figure below shows why this works: picture the remainders $0$ through $d-1$ arranged around a circle of size $d$. Adding remainders is just walking further around the circle — if you walk past the starting point, you have gone one full lap too many, so subtract $d$ (one full lap) to read off the true position.

The goal here is to prove carefully, from the division algorithm alone, that whenever two numbers share a divisor $d$, the remainder of their sum is $r_1+r_2$ if that sum is already less than $d$, and otherwise it is $r_1+r_2-d$, with the quotient increased by exactly one to compensate — never any other adjustment, however large $r_1$ and $r_2$ individually are.

Step 1. Write both numbers in division-algorithm form: $a=dq_1+r_1$ with $0\le r_1<d$, and $b=dq_2+r_2$ with $0\le r_2<d$. Adding gives $a+b = d(q_1+q_2)+(r_1+r_2)$, which is a true equation, but the term $(r_1+r_2)$ is not guaranteed to be a valid remainder yet — it only counts as the remainder of $a+b$ divided by $d$ once it satisfies $0\le (r_1+r_2)<d$.

Step 2. Bound the size of $r_1+r_2$. Since $0\le r_1<d$ and $0\le r_2<d$, adding the inequalities gives $0 \le r_1+r_2 < 2d$. So the sum of two remainders is never negative and never reaches $2d$ — it lands somewhere in the single interval $[0, 2d)$.

Step 3. Split on that interval. If $r_1+r_2<d$, the expression $a+b=d(q_1+q_2)+(r_1+r_2)$ is already in proper division-algorithm form, so the remainder is simply $r_1+r_2$. If instead $d \le r_1+r_2 < 2d$, subtract $d$ from the remainder term and add $1$ to the quotient term: $a+b = d(q_1+q_2+1) + (r_1+r_2-d)$. Because $r_1+r_2 < 2d$, the new remainder $r_1+r_2-d$ satisfies $0 \le r_1+r_2-d < d$, so this is now valid.

Step 4. Conclude. Exactly one of the two cases in Step 3 applies (the bound in Step 2 rules out needing a second subtraction), so the remainder of $a+b$ divided by $d$ is always either $r_1+r_2$ or $r_1+r_2-d$, decided purely by whether $r_1+r_2\ge d$. The same argument with a sign flip handles subtraction: if $r_1-r_2$ is negative, adding $d$ once always suffices to bring it into $[0,d)$, because $r_1-r_2 > -d$ whenever $0\le r_1,r_2<d$.

A number $a$ leaves remainder 5 on division by 8, and a number $b$ leaves remainder 6 on division by 8. What remainder does $a+b$ leave on division by 8, and what about $a-b$?

Step 1. Write $a=8q_1+5$ and $b=8q_2+6$, so $a+b = 8(q_1+q_2)+(5+6) = 8(q_1+q_2)+11$.

Step 2. Since $11\ge 8$, this is not yet valid; rewrite $11=8(1)+3$, giving $a+b = 8(q_1+q_2+1)+3$. So $a+b$ leaves remainder 3.

Step 3. Check with real numbers: $a=13$ (since $13=8\times1+5$), $b=14$ (since $14=8\times1+6$). Then $a+b=27=8\times3+3$ — remainder 3, matching. Answer: remainder 3.

Step 4. Now for the difference: $a-b = 8(q_1-q_2)+(5-6) = 8(q_1-q_2)-1$. Since $-1$ is negative, add $8$: $-1+8=7$, and subtract $1$ from the quotient part, giving $a-b=8(q_1-q_2-1)+7$. So $a-b$ leaves remainder 7. Check: $13-14=-1$; picking values where $a>b$ instead, say $a=21$ ($8\times2+5$) and $b=6$ ($8\times0+6$), gives $a-b=15=8\times1+7$ — remainder 7, matching. Answer: remainder 7.

Competitive stretch

Three numbers $x,y,z$ leave remainders 4, 7, 3 on division by 9. Find the remainder of $x+y-z$ on division by 9. Combine all three remainders at once: $4+7-3=8$, and since $0\le 8<9$ no adjustment is needed, so the remainder is 8 directly. Check: $x=13,y=16,z=12$ give $x+y-z=17=9\times1+8$. ✓

• Forgetting to subtract $d$ when $r_1+r_2$ equals or exceeds $d$ — an unadjusted sum like $11$ is not a valid remainder for divisor $8$.
• Forgetting to add $d$ when $r_1-r_2$ comes out negative, and leaving a negative "remainder" in the final answer.
• Applying the rule across different divisors — the add/subtract-remainders shortcut only works when both numbers share the same $d$.
• Assuming a second adjustment is ever needed for addition of two single remainders — since each remainder is already less than $d$, their sum is always below $2d$, so at most one subtraction of $d$ is ever required.
• Writing down an "answer" without checking it lies in $0$ to $d-1$ — a remainder of $-2$ or a remainder of $10$ for divisor $8$ is always a sign that an adjustment step was skipped.

Ask the tutor: Want to try a remainder puzzle with three or four numbers combined, or check what happens when the divisors are different for each number?