Welcome! Today we're working on Finding Missing Components in a Division Equation — a skill that will pay off big time later.

You know the division equation $a = b q + r$ and the constraint on the remainder.

Now we are going to make you fluent with this equation in every direction.

Not just dividing to find $q$ and $r$ , but also recovering any missing piece when the other three are given.

Given	Find
$b, q, r$	$a$
$a, b, r$	$q$
$a, b, q$	$r$

Why does this matter?

In Euclid's Division Algorithm, you'll chain multiple division steps
Sometimes you'll need to rearrange equations to express remainders differently

The key rearrangement:

r = a - bq

KEY

This is the exact move you will make over and over in back-substitution — the technique for writing HCF as a linear combination.

Getting this fluent now saves you from sign errors later, when the algebra gets heavier.

1. Finding the dividend from the other three components

Finding Missing Components in a Division Equation

You know the division equation $a = bq + r$ and the constraint on the remainder.

Now we're going to make you fluent with this equation in every direction — not just dividing to find $q$ and $r$ , but also working backwards!

Let's start with the simplest direction first: given the divisor, quotient, and remainder, can we reconstruct the original number?

This is the most direct application of $a = b q + r$ , and it sets the stage for the trickier rearrangements that follow.

✍️ Question

Problem 📝

A number when divided by 61 gives 27 as quotient and 32 as remainder.

What is the number?

Show your working and verify your answer.

The division equationLinks dividend, divisor, quotient, and remainder $a = b q + r$ connects four piecesKnow any three, find the fourth. When you know $b$ , $q$ , and $r$ , finding $a$ is straightforward — just compute $bq + r$ The method for finding the dividend.

Here, $b = 61$ , $q = 27$ , $r = 32$ :

a = 61 \times 27 + 32

setup

✍️ MCQ

Choose one

What is the first step to find

a

from the equation

a = 61 \times 27 + 32

?

✓

Break down $61 \times 27$ :

$61 \times 20 = 1220$ Start with the easier part of the multiplication
$61 \times 7 = 427$ Then handle the smaller multiplication

So $61 \times 27 = 1647$ answerCombine the parts to get your answer.

Add the remainder: $1647 + 32 = 1679$ .

The dividend is 1679.The reconstruction formula gives you the dividend

✍️ MCQ

Choose one

To verify our answer, we should divide

1679

by

61

. What quotient and remainder should we get if our answer is correct?

✓

⚠️ Common mistake:A frequent mistake students make Forgetting to add the remainderMust add remainder when finding dividend and writing just $1647$ .

Always remember: the dividend is $b q$ PLUSThe dividend is b times q PLUS r $r$ , not just $b q$ .

❌ IncorrectwrongMissing the remainder makes it wrong	✅ CorrectrightComplete answer includes the remainder
Dividend $= 61 \times 27 = 1647$	Dividend $= 61 \times 27 + 32 = 1679$
(Forgot the remainder!)Only multiplying is incomplete	(Includes the remainder)Includes the remainder for complete answer

✍️ MCQ

Choose one

If you forget to add the remainder when finding the dividend, your answer will be:

✓

Now verifyCheck by working backwards: divide $1679$ by $61$ Divide to see if you get original quotient and remainder.

We get $61 \times 27 = 1647$ , and $1679 - 1647 = 32$ .

The remainder is $32$ remainder32 must be less than 61, and $32 < 61$ . ✓Valid because 32 is smaller than 61

✍️ MCQ

Choose one

Why must the remainder

r

always be less than the divisor

b

?

✓

Everything checks out. ✓If quotient and remainder match, you're done

Answer: The number is $1679$ answerThe number we found.

2. Finding the divisor from the other three components

We've established that reconstructing the dividend is straightforward. Now let's tackle the reverse:

When the dividend, quotient, and remainder are known but the divisor is missing.

This requires rearranging the equation, and the verification becomes especially important because a wrong divisor might not be immediately obvious.

📋 Given Info

Given Information:

If the dividend $a$ , quotient $q$ , and remainder $r$ are known, you can find the divisor $b$ by rearranging:

a = bq + r

to get:

b = \frac{a - r}{q}

key formula

✍️ Question

Your Turn 🧮

By what number should 1365 be divided to get 31 as quotient and 32 as remainder?

Find the divisor and verify your answer by checking two conditions.

When the divisor is unknownWe set up the equation with b in place of the divisor, we need to rearrange the division equation. Starting with:

$1365 = b \times 31 + 32$

First, isolate the term with $b$ We remove the remainder by subtracting it from the dividend: $b \times 31 = 1365 - 32 = 1333$

✍️ MCQ

Choose one

We have

b \times 31 = 1333

. What operation should we perform next to find

b

?

✓

Then divide both sides by 31That's the pattern for finding any missing divisor:

b = \frac{1333}{31}

answer(The final step to find the missing divisor)

To compute $1333 \div 31$ :

$31 \times 40 = 1240$ Start by estimating with 40
$1333 - 1240 = 93$ Subtract to find the remainder
$93 \div 31 = 3$ Then divide what's left

So $31 \times 43 = 1333$ , giving us $b = 43$ answerThis two-step approach gives us the answer.

✍️ MCQ

Choose one

We found that the divisor

b = 43

. To verify this answer, we need to check two conditions. Which of the following is NOT one of those conditions?

✓

Now verify with two checks:

1. Reconstruction(Multiply divisor and quotient, add remainder, get original): Does $43 \times 31 + 32 = 1365$ ?

$43 \times 31 = 1333$ 43 times 31 gives us 1333
$1333 + 32 = 1365$ Adding 32 gets us back to 1365 ✓✓Division is correct when this works

2. Remainder constraintThe rule you must never forget: Is the remainder less than the divisor?

$32 < 43$ 32 is less than 43, so we're good ✓

Both checks pass — our answer $b = 43$ is correct!

✍️ MCQ

Choose one

Why is it necessary that the remainder must be less than the divisor?

✓

Why Both Checks Matter

Both verification checks are essential — each one catches a different type of errorEach check catches a different type of error.

The reconstruction check catches arithmetic mistakesCatches wrong multiplication errors. The remainder constraint catches logical impossibilitiesCatches things that cannot happen in division.

Example of Catching an Error

Imagine you made a calculation mistake and got divisor $= 25$ If your divisor is 25 instead of $43$ .

Check 1: $25 \times 31 + 32 = 775 + 32 = 807 \neq 1365$ ❌
Check 2: Is $32 < 25$ ? No! $32 > 25$ (FAIL)❌(A remainder of 32 is impossible with divisor 25)

The second check would immediately catch the error — a remainder of $32$ cannot come from dividing by $25$ , because the remainder must always be less than the divisorThe remainder can never exceed the divisor.

✍️ MCQ

Choose one

If someone claims that dividing a number by

20

gives a remainder of

23

, what can you immediately conclude?

✓

Key Insight

For a division equation $a = b q + r$ to be valid:

The equation must balance (arithmetic checkDoes the equation balance?): $a = b q + r$
The remainder must be less than the divisor: $r < b$ Is the remainder less than the divisor? (validity checkThe validity check for division)

Always verify both conditions!This is non-negotiable for division problems

3. Nested application of the division equation

So far we have worked with a single division equation at a time.

Now we encounter a situation where knowing the remainder from one division lets us deduce the remainder from a different division.

We are trying to find a remainder with a new divisor, using the structure hidden inside the original one.

📋 Given Info

Here's the situation:

A number when divided by $143$ gives some quotient and 31 as remainder.

So the number can be written as:

143q + 31

for some whole number

q

.

✍️ Question

Question 🤔

When this same number is divided by 13, what is the remainder?

Hint: $143 = 13 \times 11$

The number $= 143 q + 31$ for some whole number $q$ . We need the remainder when this number is divided by 13.

The key insight: 143 is a multiple of 13This is why the 143 q part gives zero remainder (since $143 = 13 \times 11$ ). So when we divide $143 q$ by 13, it goes in exactly — no remainderkey!The 143 q part divides evenly from that part.

✍️ MCQ

Choose one

If

143q

leaves no remainder when divided by 13, where does the remainder come from?

✓

The remainder must come entirely from the 31The 143 q part contributes nothing to the remainder.

Divide 31 by 13:

$13 \times 2 = 26$ , and $31 - 26 = 5$

So $31 = 13 \times 2 + 5$

This means when our original number is divided by 13, the remainder is 5answerThe equation structure reveals the answer.

✍️ MCQ

Choose one

Why did the remainder come only from

31

and not from

143q

?

✓

Putting it together:

$number = 143 q + 31 =$ $13 \times 11q$ Breaking 143 into 13 times 11 $+ ($ $13 \times 2 + 5$ Breaking 31 into 13 times 2 plus 5 $)$

Simplifying further:

$=$ $13 \times (11q + 2)$ (The crucial step is factoring out 13) $+$ (remainder) $5$ (That plus 5 is your remainder)

This is now in the form $(\text{something}) \times 13 + 5$ Something times 13, plus 5.

✍️ MCQ

Choose one

When a number is written as

13 \times (\text{something}) + 5

, what is the remainder when divided by

13

?

✓

The remainder when divided by 13 is 5The answer is 5, regardless of what $q$ is.

No matter how many times 143 went into our original number, the leftover when we divide by 13 will always be 5It's always 5 no matter what q is!

That's the power of understanding the structure of division equations!The 13 times 11 q part disappears when dividing by 13

The technique: When you need the remainder with a new divisor, check whether the original divisor is a multiple of the new oneThis multiple relationship is your green signal to use the shortcut.

If so, only the original remainder contributes to the new remainderWhen the multiple relationship exists, only the original remainder matters — apply the division equation to the remainder and the new divisor.

In our problem: Since $143 = 13 \times 11$ Dividing the quotient term by the new divisor gives zero remainder, when we divide $143 q + 31$ by 13, the $143q$ part gives remainder 0The entire quotient term vanishes because it gives zero remainder. So we only need to find the remainder when 31 is divided by 13.

✍️ MCQ

Choose one

A number gives quotient

q

and remainder 47 when divided by 91. If you want the remainder when this number is divided by 7, and you know

91 = 7 \times 13

, which value do you need to divide by 7?

✓

4. Rearranging a = bq + r to isolate the remainder

The main reconstruction skills are now in place. What remains is the specific rearrangement that will matter most going forward:

Isolating the remainder as $r = a - bq$

This form — expressing the remainder purely in terms of the dividend and divisor — is the exact move you will make at every step of back-substitution when expressing HCF as a linear combination.

✍️ Question

Given $1032 = 272 \times 3 + 216$ , express the remainder $216$ R₁ in terms of $1032$ and $272$ .

Then, given $272 = 216 \times 1 + 56$ , express $56$ R₂ in terms of $216$ and $272$ .

Why will these rearrangements be useful later?

The rearrangement from $a = bq + r$ The starting form of the division equation to $r = a - bq$ The rearranged form you need for back-substitution is simple algebra, but it's the single most important moveYou'll use this exact step every single time in back-substitution.

✍️ MCQ

Choose one

Using the equation

1032 = 272 \times 3 + 216

, express the remainder

216

in terms of

1032

and

272

.

✓

Why this matters:

When we find HCF using Euclid's algorithmGives you a chain of division equations, we get a chain of division equationsThe sequence of equations from Euclid's algorithm. To express HCF as a linear combinationWriting HCF as a x plus b y ( $HCF = ax + by$ The goal of back-substitution), we work backwardskeyYou work backwards through the chain through these equations.

Each step requires isolating the remainder: $r = a - bq$ Isolate the remainder then substitute

This lets us substitute one remainder in terms of the previous dividend and divisor, building up to the final linear combinationThat's how you build up to the final result.

✍️ FIB

Fill in the blank

From the equation

272 = 216 \times 1 + 56

, express

56

in terms of

272

and

216

.

✓

From $1032 = 272 \times 3 + 216$ , isolate 216Move other terms to the other side:

216 = 1032 - 272 \times 3

(This works for any missing term in division)

From $272 = 216 \times 1 + 56$ , isolate 56First isolate the term, then simplify:

56 = 272 - 216 \times 1</pen> = <pen actions="underline" underline-color="green" underline-style="single" narrationText="skip straight to subtraction" commentary="Just subtract directly when quotient is 1">272 - 216

(Quotient of 1 makes multiplication trivial)

✍️ MCQ

Choose one

In back-substitution, why do we need to express each remainder in terms of the dividend and divisor?

✓

Why does this matter?

When you find HCF using repeated division, you get a chain of equationsEach division gives a remainder that starts the next equation — each one connected to the next through the remaindersThe remainder feeds into what comes next.

To express the HCF as a combination of the original two numbers (like $HCF = a m + b n$ ), you work backwardsThe crucial part — backwards, not forwards through that chain.

At each step, you take a remainder and rewrite it as $r = a - bq$ Your tool at every step to isolate the remainder, then substitute into the previous expressionSubstitute into the previous expression.

✍️ MCQ

Choose one

In back-substitution, why do we need to express each remainder in terms of the dividend and divisor?

✓

If this rearrangement isn't fluent, the whole back-substitution process becomes tangledClunky rearranging now means pain later.

That's why practising these rearrangements now — like expressing $216 = 1032 - 272 \times 3$ Training you to think in r equals a minus b q form and $56 = 272 - 216 \times 1$ — builds the foundation for expressing HCF as a linear combination later.

Practise this move until it's automaticNo thinking required — it should just happen: see $a = b q + r$ , immediately know $r = a - bq$ This rearrangement finds the remainder in Euclid's algorithm.

✍️ MCQ

Choose one

Given

847 = 195 \times 4 + 67

, what is

67

equal to in terms of

847

and

195

?

✓