Welcome! Today we're tackling Picking the Divisor and Knocking Out Cases — the skills that turn the four-step template into a real problem-solving tool.

Here's the thing — most questions don't just say "list all forms."

They say things like:

"Show that every odd integer is of the form $4 m + 1$ or $4 m + 3$ "
"Show that every integer not divisible by 3 is of the form $3 m + 1$ or $3 m + 2$ "

That means you need two additional skills:

Skill	What it means
Picking the divisor	Figuring out what number to divide by
Knocking out cases	Proving why certain remainders can be thrown out

Let's build both.

By the end of this section, you'll know exactly how to choose your divisor and eliminate the cases that don't apply.

1. Reading the divisor from the target form

The proof template says 'choose a divisor' as step 1 — but how do you know which divisor to choose?

It turns out the question itself tells you, if you know where to lookkey.

📋 Given Info

When a question says 'show that every positive integer is of the form $3 m$ , $3 m + 1$ , or $3 m + 2$ ,' the forms all involve $3m$ .

When a question says 'show that every positive odd integer is of the form $4 m + 1$ or $4 m + 3$ ,' the forms involve $4m$ .

Key insight: The coefficient of $m$ in the given forms tells you the divisor to use.

✍️ Question

A question asks you to show that certain integers are of the form $6m + 1$ , $6m + 3$ , or $6m + 5$ .

What divisor should you use in the division lemma, and how many total cases will you need to list before you start eliminating?

Here's the rule for picking the divisor: look at the coefficient of $m$ This coefficient becomes your divisor (or $q$ , or whatever variable the question uses) in the target form.

Target form involves $3 m + . . .$ → divide by 3
Target form involves $4 m + . . .$ → divide by 4
Target form involves $6 m + . . .$ → divide by 6÷6The coefficient tells you exactly what to do

The divisor is always this coefficient.

✍️ MCQ

Choose one

When you divide by

6

using the division lemma, the possible remainders are

0, 1, 2, 3, 4, 5

. How many total cases do you start with before eliminating any?

✓

Why does this work? Because when you write $n = b m + r$ , the number $b$ The coefficient b equals the divisor in front of $m$ is literally the divisorThey're the same thing, always in the division lemma. It's not a coincidence — it's built into the structure of the equationIt's how the equation is structured!

For the question about $6 m + 1$ , $6 m + 3$ , $6 m + 5$ : the coefficient is 6The coefficient tells you how many cases, so divide by 6divisorDivide by 6. This gives 6 total casesSix cases to start with — one for each remainder $0, 1, 2, 3, 4, 5$ Your starting point before elimination.

✍️ MCQ

Choose one

If a question asks you to show that certain integers are of the form

5m + 2

or

5m + 4

, what divisor would you use?

✓

⚠️ A common mistake:Don't fall for this common error thinking the divisor is the number of target forms (here, 3wrong!Counting forms gives the wrong divisor because there are three forms listed).

But the divisor comes from the coefficientLook at the coefficient, not the count, not the count.

In forms like $6 m + 1$ , $6 m + 3$ , $6 m + 5$ , the coefficient of $m$ is 6divisorSix is our divisor here — so the divisor is 6, giving us 6 casesRemainders 0 through 5 are our cases to start with (remainders 0, 1, 2, 3, 4, 5).

✍️ MCQ

Choose one

A question asks you to show integers are of the form

5m + 1

or

5m + 4

. How many total cases do you start with before eliminating?

✓

2. Eliminating cases with algebraic justification

Good — you can pick the right divisor.

But the question only asks you to show three of the six forms are valid for odd integers.

That means three forms need to be eliminated, and 'it's obviously even' isn't enough. You need algebra.

📋 Given Info

Here's where we are:

You're proving: 'Every positive odd integer is of the form $4 m + 1$ or $4 m + 3$ .'

You've applied the division lemma with divisor 4 and listed four forms:

$4 m$
$4 m + 1$
$4 m + 2$
$4 m + 3$

✍️ Question

Your task ✍️

Which of the four forms must be eliminated, and why?

For each form you eliminate, write the algebraic factorisation that proves it is even.

To prove a form is even, you need to show it can be written as $2 \times$ (something)The test for evenness — can you factor out a 2?.

That's your test — if you can factor out a $2$ , it's evenThis is how you prove evenness in any algebra problem. If you can't, it's odd.

Let's check each of the four forms systematically:

$4m = ?$
$4m + 1 = ?$
$4m + 2 = ?$
$4m + 3 = ?$

✍️ FIB

Fill in the blank

Look at

4m

. Can you write it as

2 \times

(something)? If yes, what is that something?

✓

4m: Is this even?

$4 m = 2 \times 2 m = 2 (2 m)$

Yes — it's 2 times $(2m)$ This is your test for evenness. Even. Eliminate.out4m has a factor of 2, so it's out

4m + 1: Can we write this as 2 times something?

$4 m + 1 = 2 (2 m) + 1$

The leftover 1Can't divide evenly by 2 means it's odd. Keep.

✍️ MCQ

Choose one

What is the algebraic factorisation that proves

4m + 2

is even?

✓

4m + 2: Is this even?

$4 m + 2 = 2 (2 m + 1)$

Yes — it's 2 times $(2m + 1)$ Every term has the same factor. EvenThat's how you spot even numbers quickly. Eliminate.

4m + 3:

$4 m + 3 = 2 (2 m + 1) + 1$

OddSame logic as before. Keep.✓The only forms that can be odd

✍️ MCQ

Choose one

After eliminating the even forms, which two forms represent all positive odd integers?

✓

So we eliminate $4 m$ and $4 m + 2$ , keeping $4m + 1$ and $4m + 3$ .

The conclusion: Since every integer is one of these four forms (by the division lemmaGuarantees we've covered all possibilities) and two are even, every ODDThe only remaining options after eliminating even cases integer must be one of the remaining two:

4m + 1 \quad \text{or} \quad 4m + 3

answer(The only forms an odd integer can take)

✍️ MCQ

Choose one

Using the same approach, if we divide by

6

, we get six forms:

6m

,

6m+1

,

6m+2

,

6m+3

,

6m+4

,

6m+5

. How many of these forms represent odd integers?

✓

The algebraic factorisation is what earns the marks.

"Clearly 4m is even" is an assertionClaiming something is true without showing why.
" $4 m = 2 (2 m)$ " is a proofShows WHY by writing it as 2 times something.

✍️ MCQ

Choose one

Which of the following is a valid proof that

4m + 2

is even?

✓

3. Full elimination proof with divisor 6

You can eliminate with divisor 4. Now let's scale up — more cases to manage, more factorisations to write.

The logic is identical, but the bookkeeping gets more involved.

📋 Given Info

Consider the claim: 'Every positive odd integer is of the form $6m + 1$ , $6m + 3$ , or $6m + 5$ .'

You've chosen divisor 6 and listed all six forms:

$6m$
$6 m + 1$
$6m + 2$
$6 m + 3$
$6m + 4$
$6 m + 5$

✍️ Question

Complete the proof.

For each form you eliminate, show the factorisation.

For each form you keep, briefly explain why it's oddgoal.

Let's work through all six forms systematically. The test: can you write the form as $2 \times$ (a whole number)Just check if it equals 2 times a whole number?

$6m = 2(3m)$ . Even — the factor of 2 is clear. Eliminate.

$6 m + 1 = 2 (3 m) + 1$ . You can pull out a 2 from the $6 m$ part, but you're left with a $+1$ leftoverThis plus 1 won't divide evenly by 2 that won't divide by 2. So this is odd. Keep.

✍️ MCQ

Choose one

What is

6m + 2

written as

2 \times

(something)?

✓

6m + 2 = 2(3m + 1)Factor out 2 to check if even. EvenDivisible by 2 means even — factor the whole thing by 2. Eliminate.Even numbers get eliminated

6m + 3 = 2(3m + 1) + 1The remainder after factoring by 2. The 6m + 2 part is even, and adding 1Adding 1 to an even number makes it oddEven plus 1 equals odd. Keep.Odd numbers we keep

✍️ MCQ

Choose one

What is the factorisation that proves

6m + 4

is even?

✓

6m + 4 = 2(3m + 2)Same factoring pattern as before. EvenEven means divisible by 2. Eliminate.Even forms get eliminated

6m + 5 = 2(3m + 2) + 1Remainder of 1 after factoring. OddOdd forms remain as candidates. Keep.Odd forms we keep

Pattern: Every even remainder (0, 2, 4)Even remainders produce even expressions gives an even form, and every odd remainder (1, 3, 5)Odd remainders produce odd expressions gives an odd form.

Makes sense — adding an even number to an even number ( $6m$ Six times anything is always even) keeps it even, and adding an odd number makes it odd.

✍️ MCQ

Choose one

If you were proving that every positive odd integer is of the form

8m + 1

,

8m + 3

,

8m + 5

, or

8m + 7

, which remainders would you eliminate?

✓

Conclusion: Three forms eliminated(Forms with even remainders are always divisible by 2) ( $6 m$ , $6 m + 2$ , $6 m + 4$ ), three keptOdd remainders mean they can actually be odd numbers ( $6 m + 1$ , $6 m + 3$ , $6 m + 5$ ).

Since the division lemmaGuarantees there are no other possibilities guarantees every integer is one of the six forms, and three of them are even, all odd integers must be among the remaining threeThere's nowhere else for odd numbers to go: $6 m + 1$ , $6 m + 3$ , or $6 m + 5$ .

✍️ MCQ

Choose one

In the proof we just completed, what did we use the division lemma to guarantee?

✓

4. Avoiding the wrong starting point

There's a common detour students take on these problems — one that's technically correct but makes the proof much harder than it needs to be.

Recognising this trap saves time and frustration.

📋 Given Info

Here's the scenario:

A student sees the question:

'Show that every positive odd integer is of the form $4 m + 1$ or $4 m + 3$ .'

They immediately write:

n = 2k + 1 \quad \text{(since } n \text{ is odd)}

...and then try to work from there to get forms involving $4m$ .

✍️ Question

Think about this approach 🤔

What's wrong with this approach? Will it eventually work? And why is dividing by 4 directly a better starting point?

Starting with $n = 2k + 1$ This form is valid but inefficient isn't wrong — every odd number can be written this way. The problem is that it doesn't get you to forms involving $4 m$ . You'd need a second division stepForces extra work we want to avoid.

Here's what happens: you write $n = 2k + 1$ Starting with 2k plus 1 leads to more steps, and then you realise you need to know what form $k$ takesYou end up asking this extra question. So you apply the division lemma to $k$ with divisor 2: $k = 2j + r$ Two applications instead of one where $r = 0$ or $1$ .

✍️ MCQ

Choose one

If you had directly applied the division lemma to

n

with divisor

4

, how many division steps would you need?

✓

If $k = 2 j$ : $n = 2(2j) + 1 = 4j + 1$

If $k = 2 j + 1$ : $n = 2 (2 j + 1) + 1 = 4 j + 2 + 1 =$ $4j + 3$

✍️ MCQ

Choose one

In the approach starting with

n = 2k + 1

, how many times did we apply the division lemma to reach the final forms

4j + 1

and

4j + 3

?

✓

You get the right answer, but through two applicationsWe needed two steps instead of one of the lemma.

Compare this to dividing by 4 directly: one stepDirect division saves effort gives you $4 m$ , $4 m + 1$ , $4 m + 2$ , $4 m + 3$ , and you eliminate the even ones. Same result, half the work.

The takeaway: When a problem asks for forms involving a specific number (like $4 m$ ), use that number as your divisor right from the startIf it says 4m, divide by 4 from the start. Don't take a detour through $2k + 1$ and then split againExtra splitting is unnecessary work — go straight to the source!

The lesson: when the target form involves $b m$ , divide by $b$ straight away.

Don't start with a smaller divisor and work up — you're just making extra work for yourself.

Rule of thumb: Look at the question. See $4m$ ? Divide by 4.Divide by 4 when you see 4m in the question See $6 m$ ? Divide by 6. Match the divisor to the target form.Your shortcut to avoid extra steps

✍️ MCQ

Choose one

If a problem asks you to prove something about forms like

5m

,

5m + 1

,

5m + 2

, etc., what should you divide by?

✓