Notebook

00:06

30 Mar 2026

Subtract and Find HCF: When Remainders Are Given

Welcome! Today we're going to master a specific technique for finding the Highest Common Factor in a new scenario: Subtract and Find HCF: When Remainders Are Given.

You already know how to compute HCF when numbers divide exactly.

But what if division leaves a remainder?

In this section, you will discover a single algebraic move — subtracting the remainder — that converts any remainder problem into one you already know how to solve.

We will apply this technique to several scenarios:

Problems involving two and three numbers
Cases where adjusted numbers have no obvious factors
Situations requiring systematic prime testing

This is one of six patternskey! for remainder-adjusted problems you'll learn. Outside of exams, this reasoning helps find a common measure in real-world tasks:

Example	Scenario
Manufacturing	Cutting rods of different lengths into equal pieces
Logistics	Organizing items with specified leftover amounts

1. The remainder-to-exact-division conversion

You know how to find HCF when numbers divide exactly. But what happens when division leaves a remainder?

Before we can solve these problems, we need a way to convert:

'divides with remainder' $\to$ 'divides exactly'

There is a simple algebraic move that does this.

📋 Given Info

Here's the situation:

We have a number 545
When we divide it by some number $d$ , we get a remainder of 5extra

Goal: We want to find the value of $d$ .

✍️ Question

Think about this: 🤔

If 545 leaves a remainder of 5 when divided by $d$ , what number does $d$ divide exactly (with no remainder)?

Why?

The Fundamental Rule of DivisionYour starting point for remainder problems

When a number $N$ leaves a remainder $r$ upon division by a divisor $d$ , it can be expressed as:

$N = (d \times q) + r$

Where:

$N$ : The original number (Dividend)
$d$ : The divisor
$q$ : The quotient (how many times $d$ fits into $N$ )
$r$ : The remainderThe amount that prevents perfect division — the "extra bit" that doesn't fit into a perfect group of $d$ .

✍️ MCQ

Choose one

If division is perfect, what is the value of

r

Making it Perfectly DivisibleConverts to a factor problem you can solve

By rearranging the division equation, we get a powerful tool for solving HCF problems:

$N - r = d \times q$

Key Insight: If you subtract the remainderImmediately make the number perfectly divisible from the original number, the resulting value is perfectly divisibleThe key trick when a problem mentions a remainder by the divisor $d$ .

To get rid of a remainder, simply subtract it from the original number!Subtract it to solve using HCF

✍️ MCQ

Choose one

23

leaves remainder

3

, which number is exactly divisible by

d

Application Example

If we have the number 545We don't work with this directly because of the remainder and it leaves a remainder of 5This prevents direct division when divided by $d$ :

Subtract the remainder: $545 - 5 = 540$
Identify the relationship: $540 = d \times q$
Conclusion: The divisor $d$ must divide 540resultYour mystery divisor must be an exact factor of this exactly.

This confirms that our mystery divisor is a factor of the adjusted numberThe divisor must be an exact factor of this new value.

The Core Strategy

This single insight — subtracting the remainder converts a remainder problem into an exact-division problemThis converts messy division into exact division — is the engine behind the entire technique.

By simply subtracting the remainderAlways subtract first to clean up your numbers, we "clean up" the numbers before starting the real work. This removes the messiness of remainders, making the problem much simpler to solve.

✍️ MCQ

Choose one

Which number is exactly divisible by the divisor if

103

leaves a remainder of

3

Turning Remainders into HCF

Instead of looking for "the largest $d$ that divides $545$ with remainder $5$ ," you simply look for "the largest $d$ that divides $540$ exactlyFind a number that divides exactly."

Why does this work?

Subtracting the remainder ( $545 - 5 = 540$ ) creates a number that has no remainder when divided.
The "largest number" that divides others exactly is, by definition, the Highest Common Factor (HCF)This is the definition of H C F.

Rule: Every remainder problemkey insightExam remainder problems are H C F in disguise is just an HCF problem in disguiseIt's an H C F problem after the subtraction once you subtract those remainders!

✍️ Yes/No

Yes or No?

Does the word 'largest' in the phrase 'largest divisor' suggest that we should find the HCF?

2. Applying the subtraction method with two numbers

We have the key insight:

Subtracting the remainder turns a remainder problem into an exact-division problem.

Now we need to put this to work on a full problem:

Adjust both numbers by subtracting their remainders
Factorise the new numbers
Compute the HCF ( $H i g h e s t C o m m o n F a c t o r$ )

📋 Given Info

Here's the problem:

You want to find the largest number that divides 129 and 545, leaving remainders 3 and 5 respectively.

The Key Strategy: Subtracting each remainder from its corresponding number gives you values that the required number divides exactly.

$129 - 3 = 126$
$545 - 5 = 540$

✍️ Question

Your Turn ✏️

Find the largest number that divides 129 and 545, leaving remainders 3 and 5 respectively.

Show all steps:

Adjustment (subtract the remainders)
Factorisation (prime factorise the adjusted numbers)
HCF (find the highest common factor)

Problem Setup: Subtract and Find HCF

Find the largest numberThis signals we need to find HCF that divides 129 (leaving remainder 3) and 545 (leaving remainder 5).

Step 1 — Adjust the Numbers

To find a number that divides perfectly, we must first "remove" the extra remainder from our starting values.

For 129: $129 - 3 = 126$ adjustedRemove the leftover to get exact division

Note: Our mystery number must divide 126 exactlyNo remainder means perfect division, without any remainder.

✍️ FIB

Fill in the blank

What is

545 - 5

Step 2 — Adjust the Second Number

For 545: $545 - 5 = 540$ adjustedThe second number for your calculation

Our mystery number must also divide 540 exactly.

⚠️ Key Rule: Match the Remainders

Notice how we paired them:

$129 \rightarrow 3$ Each remainder pairs with its own number
$545 \rightarrow 5$ Each remainder pairs with its own number

Each number gets its OWN specific remainder subtracted. This is crucial for accuracy, especially when the remainders are different. Do not mix them up!A common mistake to avoid

✍️ MCQ

Choose one

Which is the correct adjustment for

Y

Step 2: Prime Factorization

Now that we have our adjusted numbers, 126 and 540, we need to find their prime factors to move forward. Let's break them down systematically.

Factoring 126

Divisible by 2? Yes, it ends in an even number. $126 = 2 \times 63$ Always begin with 2 when the number is even.
Divisible by 3? Check the sum of digits: $6 + 3 = 9$ . Since 9 is a multiple of 3, 63 is divisible by 3.
Breaking it down: $63 = 3 \times 21$ , and $21 = 3 \times 7$ .

Prime Factorization of 126:

126 = 2 \times 3^2 \times 7

Result(This is the foundation for finding the HCF)

✍️ FIB

Fill in the blank

In the prime factorization of

126 = 2 \times 3^n \times 7

, what is the value of

n

Factoring 540

Next, let's tackle the larger number, 540For bigger numbers, be systematic with your division, by dividing by prime factors step-by-step:

Divide by 2: It's even, so $540 = 2 \times 270$ Keep dividing by 2 until you can't anymore.
Divide by 2 again: $270 = 2 \times 135$ .
Divisible by 3? Check the sum of digits for 135: $1 + 3 + 5 = 9$ Add the digits: 1+3+5=9, which is a multiple of 3. This means it is definitely divisible by 3When digit sum is divisible by 3, the number is too.
Breaking down 135: $135 = 3 \times 45$ .
Further factors: $45 = 3 \times 15$ , and $15 = 3 \times 5$ .

Prime Factorization of 540: $540 = 2^{2} \times 3^{3} \times 5$

✍️ MCQ

Choose one

Why is

135

divisible by

3

Step 3 — Finding the HCF

To find the Highest Common Factor (HCF) from prime factorizations, we follow the Golden Rule:

The Rule

Take the lowest powerThe largest amount both numbers share of each COMMON prime factorMust appear in both lists.

Note: Only include prime numbers that appear in the factorization of both numbers. If a factor is missing from one listSkip factors not in both, it is not common and cannot be part of the HCF.

✍️ MCQ

Choose one

Which prime factors are common to both

126

and

540

Comparing Common Factors

Identify the common primes and compare their exponents to find the smallest power:

Prime 2: Appears in both lists.
- Powers: $2^{1}$ and $2^{2}$
- Smallest power (min) = $2^1$ winnerOne is smaller than two
Prime 3: Appears in both lists.
- Powers: $3^{2}$ and $3^{3}$
- Smallest power (min) = $3^2$ winnerTwo is smaller than three

✍️ MCQ

Choose one

Why do we skip the prime factor

7

in our HCF calculation?

Final Calculation

We exclude any factors that are not shared:

Prime 7: Only in $126$ . Not commonOnly appears in one list $\to$ Skip.
Prime 5: Only in $540$ . Not commonCannot be part of the HCF $\to$ Skip.

Multiply the "winners"Multiply the lowest powers together (the lowest powers of common primes) to get the result:

$HCF = 2^{1} \times 3^{2}$

\text{HCF} = 2 \times 9 = 18

(The final HCF result)

Conclusion: $18$ answerDivides both numbers evenly is the largest number that divides our original values with the specified remainders.

Verification: Why it Works

It's always a good idea to check our work! Let's verifyChecking our work catches errors if 18 is truly the largest number that divides 129 and 545 while leaving the required remainders.

The Goal:

Divide 129 by 18 $\to$ Should leave a remainder of 3targetMust match the remainders from the problem.
Divide 545 by 18 $\to$ Should leave a remainder of 5.

In math, we don't just take our own word for it—let's prove it by dividingDividing originals by HCF proves correctness!

✍️ FIB

Fill in the blank

What is the remainder when

129

is divided by

18

Verification Part 1: 129

$129 \div 18 = 7$ with a remainder of 3Three matches what was required for 129 ✓

Since $18 \times 7 = 126$ , subtracting that from 129 leaves exactly 3. That matches our requirement perfectly!

Now, let's test the second number: 545Testing if 18 works for the second number too.

✍️ MCQ

Choose one

18 \times 30 = 540

, what is the remainder when

545

is divided by

18

Verification Part 2: 545

$545 \div 18 = 30$ with a remainder of 5 ✓

Since $18 \times 30 = 540$ , subtracting that from 545 leaves a remainder of exactly 5.

Conclusion: Both remainders match the original question perfectly! This proves that 18answer18 is the largest number satisfying both conditions is the correct HCF for our adjusted numbers and the correct answer for this problem.

Great work! We've successfully used the Subtract-then-HCFSubtract remainders first, then find HCF of results method.

3. Systematic factorisation of unfamiliar numbers

We have the adjustment and HCF technique working for numbers like $126$ and $540$ , which have obvious small factors.

But many remainder-adjusted problems produce numbers like 391 or 527:

Numbers that do not divide by $2$ , $3$ , or $5$ .
The factorisation step becomes the bottleneck.

Here's the situation:

📋 Given Info

You need to find HCF(391, 425, 527). These are adjusted values from a three-number remainder problem.

425: Ends in $5$ , so it is clearly divisible by $5$ .
391 and 527:
- Not even (not divisible by $2$ )
- Sum of digits is not a multiple of $3$
- Do not end in $0$ or $5$

When small prime factors like $2, 3,$ and $5$ fail, we need a systematic approach to factorise.

✍️ Question

Factorise 391 into its prime factors.

Show which primes you test
Show how you check each one for divisibility

Testing Factors Systematically

Numbers like 391This number trips up many students in exams can be tricky because they do not have small, obvious factors. At first glance, it doesn't immediately jump out as being in any simple multiplication table.

Why 391 is Tricky:

Not Even:It doesn't end in an even digit It doesn't end in an even digit, so it's not divisible by $2$ .
No 5s or 0s:So we know it's not divisible by 5 It doesn't end in $0$ or $5$ , so it's not divisible by $5$ .
Digit Sum Rule:The sum of digits isn't a multiple of 3 The sum of the digits ( $3 + 9 + 1 = 13$ ) is not a multiple of $3$ , so it's not divisible by $3$ .

When we are stuck like this, we don't guess randomly. We need to test prime numbers systematicallyOtherwise you'll waste time guessing to uncover hidden factors.

✍️ Yes/No

Yes or No?

391

divisible by

2

5

The Square Root Rule

How far do we actually need to test? Do we have to check every prime all the way up to $391$ ? No!

The Rule: To find the factors of a number, you only need to test prime numbers up to the square rootGives you a clear stopping point for testing of that number.

For our number:

\sqrt{391} \approx 19.8

≈19.8(Any factor must be 19 or smaller)

The Plan: Since the square root is approximately 19.8, this gives us a clear stopping pointWe know exactly where to stop testing. You only need to test primes up to 19. This "secret"Prevents checking every number up to 391 saves an immense amount of time and effort!

✍️ MCQ

Choose one

What is the largest prime number less than or equal to

19.8

Our Systematic Checklist

Based on the Square Root Rule, our list of potential prime factors is much smaller. Instead of checking hundreds of possibilities, we only need to test:

2, 3, 5, 7, 11, 13, 17, 19These are the only ones we need to check

Why this works:

Efficiency:Our checklist is dramatically reduced We've narrowed the search down to just these few numbers.
The Prime Test: If none of these primes divide $391$ exactly, then we have mathematically proven that $391$ is a prime numberprime!Gives you confidence to stop searching itself!

Let's test these one by one to find the hidden factorsThe factors that aren't immediately obvious.

Testing the Primes: 2, 3, and 5

To find the factors of 391, we'll start our checklist with the easiest prime numbersStart with the simplest primes to save time. Since it's a large number, we'll work through the list systematically:

Divisibility by 2: Look at the last digit ( $1$ )Check if the last digit is odd or even. It is an odd numberOdd numbers can't be divided by 2.
- Conclusion: 391 is not divisible by 2.
Divisibility by 3: Sum of the digits: $3 + 9 + 1 = 13$ Add the digits together to test for 3.
- Observation: Since 13 is not in the 3 times table13 is not in the 3 times table (not divisible by 3), 391 is not divisible by 3.
Divisibility by 5: Check the last digit. It is not a $0$ $0$ or a $5$ $5$ .
- Conclusion: 391 is not divisible by 5.

Moving through the list quickly!

✍️ MCQ

Choose one

Why is

391

not divisible by

3

Testing the Primes: 7, 11, and 13

Next, we test the primes that require a bit more division and calculationDivision is needed for primes like 7, 11, 13:

Try 7:
- $7 \times 55 = 385$ .
- $391 - 385 = 6$ .
- Result: Since the remainder is $6$ remainderA remainder means it's not a factor (not zero), 7 does not divide 391.
Try 11:
- $11 \times 35 = 385$ .
- If we add another 11, we get $396$ (too big).
- Result: 391 is not divisible by 11.
Try 13:
- $13 \times 30 = 390$ .
- The difference is only $1$ !
- Result: Close doesn't count for factorsEven a remainder of 1 means no perfect division. 13 is not a factor.

✍️ FIB

Fill in the blank

What is the remainder when

391

is divided by

13

The Breakthrough: Testing 17

Don't give up!Keep testing larger primes like 17 We have one more prime from our list to test.

Try 17:
- $17 \times 23 = 391$ Sometimes the factor is a bigger prime.

The division is exact! ✓exact!Exact division means you found a factor

We finally found it! This means $391$ isn't prime after allIt can be factored into primes. It is the product of two prime factors.

Prime Factorisation

$391 = 17 \times 23$ (391 equals 17 times 23)

This was the hidden factor we needed to continue our HCF calculation.

⚠️ The "Small Prime" Pitfall

Many students make the mistake of testing only the smallest prime numbers like 2, 3, and 5Testing only these small primes is not enough. When these don't work, they wrongly conclude the number is prime.

Don't let easy numbers fool you! As seen with $391$ exampleLooks prime but is actually composite, a number can appear prime at first glance but actually have much larger prime factors.

✍️ Yes/No

Yes or No?

If a number is not divisible by

2, 3

, or

5

, does that guarantee it is a prime number?

🔍 Persistence in Testing

When dealing with larger numbers (specifically in the 300-600 rangeMust be persistent with division in this range), you cannot stop early. You must be persistentKeep testing, don't stop early in your search for factors.

Rule of Thumb:

Test all prime numbers sequentially.
You must test up to at least 19 or 23Keep testing until you reach these primes to be certain of the number's primality.

✍️ MCQ

Choose one

Why should we test up to

23

for a number like

529

Systematic Factorization

Some numbers don't show their factors easily, so it is important to be persistent and test prime numbers systematicallyDon't give up if easy primes don't work.

Testing 17: By dividing $527$ by $17$ , we find it goes exactly $31$ times.
Factorization: $527 = 17 \times 31$ (Recognizing larger prime factors is the hardest part)

Both $17$ and $31$ are prime numbers, so this is the complete prime factorization.

✍️ MCQ

Choose one

17

a prime number or a composite number?

Prime Factorization of 425

Since $425$ ends in $5$ Use this to simplify your work immediately, we can easily identify $5$ as a factor:

$425 = 5 \times 85$
Further breaking down $85$ : $85 = 5 \times 17$
Final Form: $425 = 5 \times 5 \times 17 = 5^2 \times 17$ Use powers to keep factors organized for HCF

Observation: Notice that $17$ common!Spotting repeating primes is your biggest clue appears here just as it did in the factorization of $527$ .

✍️ MCQ

Choose one

Which prime factor is common to both

527

and

425

Final HCF Calculation

By comparing the prime factorizations of all three numbers in our set:

$391 = 17 \times 23$
$425 = 5^{2} \times 17$
$527 = 17 \times 31$

We identify that $17$ shared!The key step is identifying the shared prime is the only prime factor shared by all three numbers.

Result:

\text{HCF}(391, 425, 527) = 17

(If a number doesn't appear in every list, it can't be in the HCF)

4. Handling the same-value remainder (both remainders are equal but given)

We've covered how to adjust, factorise, and compute HCF when remainders are given. Now let's look at one last subtlety:

What if the problem states the same remainder for both numbers?

This edge case sits right at the boundary between the technique you've learned (subtracting given remainders) and the pairwise-difference technique you'll learn next.

📋 Given Info

Consider this problem:

"Find the largest number that divides 438 and 606, leaving remainder 6same! in each case."

Key Observation:

Both remainders are 6 — they are the same value.
This value is given explicitly in the problem statement.

✍️ Question

Your turn 🤔

How do you solve this problem?

Do you subtract the remainders (like the given-remainder pattern)?
Or do you take pairwise differences (like the unknown-remainder pattern)?

Explain why, and find the answer.

A Subtle but Important Distinction

Key Insight: The phrasing 'leaving remainder 6 in each case'The remainder is already known for both numbers is crucial.

Because the remainder value is explicitly givenThe remainder is stated directly in the problem to us as 6 for both numbers, we don't need to find it ourselves. We can move straight to the adjustment stepMake the numbers divisible right away to transform this into an exact-division problem.

✍️ MCQ

Choose one

Since the remainder is given as

6

, what is our first step?

The 'Given Remainder' Method

To find the divisor that leaves a remainder, we first subtract the given remainder from each original numberDo this before anything else to find the adjusted numbers:

Number 1: $438 - 6 = 432$
Number 2: $606 - 6 = 600$

These adjusted numbers ( $432$ and $600$ ) are now values that our target divisor will divide exactlykeyThe new values become a standard HCF task.

✍️ FIB

Fill in the blank

The adjusted numbers are:

Finding the HCF

Now that we have our adjusted numbers, the final task is clear. To find the largest number that divides both $432$ and $600$ exactly, we calculate:

\text{HCF}(432, 600)

(Calculate HCF of the adjusted numbers)

This HCF will be the largest number that leaves a remainder of $6$ This is your final answer when dividing the original numbers $438$ and $606$ .

Step 2: Factorise our adjusted numbers

Now that we have our adjusted numbers, we need to find their prime building blocksThe prime factors that multiply to make the number through prime factorisation.

Factorise 432: Since 432 is an even number, we repeatedly divide by 2 until we reach an odd number.

$432 = 2 \times 216$
$216 = 2 \times 108$
$108 = 2 \times 54$
$54 = 2 \times 27$
$27 = 3^3$ 3³Recognize this as three cubed

Final Prime Factorisation: $432 = 2^{4} \times 3^{3}$

✍️ FIB

Fill in the blank

What is

3^3

Factorise 600: Similarly, we break down 600 into its prime factors. We divide by 2 three times, then move to the next prime factors, 3 and 5.

$600 = 2 \times 300$
$300 = 2 \times 150$
$150 = 2 \times 75$ An odd number so we stop dividing by two
$75 = 3 \times 25$
$25 = 5^2$ 5²Recognize it as five squared

Final Prime Factorisation: $600 = 2^{3} \times 3^{1} \times 5^{2}$

✍️ Yes/No

Yes or No?

Is the prime factor

5

common to both

432

and

600

Step 3: Find the HCF

To find the Highest Common Factor (HCF)The highest common factor we need to find, we follow this rule:

Rule: Take the lowest power of each COMMON prime factor.Only look at shared primes with lowest exponents

Let's compare the powers in our factorisations:

Prime 2: We have $2^{4}$ and $2^{3}$ . The smaller power is 3. $\rightarrow 2^3$ ✓Pick the smaller power for each shared base
Prime 3: We have $3^{3}$ and $3^{1}$ . The smaller power is 1. $\to 3^{1}$
Prime 5: This is not common to both numbers, so we ignore itFive isn't in both numbers so skip it.

Calculation:

\text{HCF} = 2^3 \times 3 = 8 \times 3 = 24

(Multiply the common primes with lowest powers)

Thus, the HCF of our adjusted numbers is 24.

Verify the Results

It is always a good practice to check our workThe only way to be certain your HCF is correct to ensure the calculated HCF satisfies the original conditions.

Step 1: Check the first number

Division: $438 \div 24 = 18$ with a remainder of $6$ ✓Must match exactly what the question asked for
Verification: Since the problem specifically asked for a remainder of $6$ , this result is correct ✓

✍️ FIB

Fill in the blank

24 \times 25 = 600

, what is the remainder when

606

is divided by

24

Step 2: Check the second number

Division: $606 \div 24 = 25$ with a remainder of $6$ ✓Same remainder confirms your HCF works ✓

Conclusion: Both numbers, when divided by $24$ , leave the required remainder of $6$ . Our answer is officially verified!You can move on with full confidence

Great job following through the steps!

Distinguishing Remainder Methods

It is very important not to mix up our methods!A mistake many students make The strategy you choose depends entirely on what the problem tells you about the remainder.

1. The Given Remainder Method(Your go-to when an exact value is provided)

If the problem provides an exact value (e.g., "leaves a remainder of 6"), you simply subtract that specific value from each of your original numbers first.

2. The 'Unknown Same Remainder' Method

This method (using pairwise differences) is used ONLY when the problem says "leaves the same remainder" WITHOUT telling you what that value is.

Key Logic: You cannot subtract a number you don't know!StopYou must switch your strategy entirely If the value is a mystery, you must subtract the original numbers from each other instead.

✍️ MCQ

Choose one

If a problem states: "Find the largest number that divides

100

and

150

leaving a remainder of

4

in each case," what is your first step?

Compare the Phrasing:

Use this table to quickly identify which action to take based on the language of the question:

Phrasing	Remainder Value	Action
"Leaving remainder 6 in each case"	GIVEN (e.g., 6)	Subtract the valueWhen the value is given, subtract it from each number.
"Leaving the same remainder in each case"	NOT GIVEN (Unknown)	Take pairwise differencesFor unknown values, subtract numbers from each other (subtract the numbers from each other).

Summary Tip:

Given value? Subtract that value.
Unknown value? Subtract the numbers from each other.

✍️ T/F

True or False?

When the remainder value is NOT provided in the problem, we use the 'pairwise differences' method.