Welcome! Today we're tackling Remainder-Adjusted HCF — the technique you need when numbers don't divide cleanly.

You've just learned to identify HCF problems from signal words. Now here's a twist: what if the problem says 'find the largest number that divides 245 and 1037, leaving remainder 5 in each case'?

The numbers don't divide cleanly — there's a leftover.

If you compute HCF(245, 1037) directly, you get the wrong answer.

There's one extra step you need — and it's where most students lose marks.

By the end of this section, you'll handle:

Type	Description
1	Same remainder given
2	Different remainders given
3	Remainder unknown

Plus — the elegant trick for when the remainder itself is a mystery.

1. Why you subtract the remainder before finding HCF

Remainder-Adjusted HCF: Subtract Before You Compute

You've just learned to identify HCF problems from signal words. Now here's a twist:

What if the problem says 'find the largest number that divides 245 and 1037, leaving remainder 5 in each case'?

The core insight of remainder-adjusted problems is a single algebraic move. Once you see why it works, all three problem variants become straightforward.

📋 Given Info

Given Information:

A problem says: 'Find the largest number that divides 245, leaving remainder 5.'

This means when you divide 245 by the mystery number $N$ , you get remainder 5.

In other words: $245 = N \times q + 5$ (where $q$ is some quotient)

This is the division algorithm: Dividend = Divisor × Quotient + Remainder

✍️ Question

Question 🤔

If $N$ divides 245 leaving remainder 5, what number does $N$ divide EXACTLY (with no remainder)?

Write the algebraic equation and explain why subtracting the remainder converts an inexact division into an exact one.

Let's trace the algebra carefully.

The problem says: N divides 245 leaving remainder 5Write the division equation from this statement. This means:

245 = N \times q + 5

(The standard division equation format)

\text{ (for some quotient } q\text{)}

Rearrange: $245 - 5 = N \times q$

240 = N \times q

key result(N divides 240 exactly, no remainder left)

So N divides 240 exactlyA clean multiple of the divisor.

✍️ MCQ

Choose one

If a number

N

divides

378

leaving remainder

8

, what number does

N

divide exactly?

✓

The remainder 5 is the 'extra' part that prevents 245 from being a clean multiple of N. Removing it (subtracting 5Subtract remainder to get clean division) gives you a number that IS a clean multiple.

Key insight: Subtracting the remainder converts an inexact division into an exact one!The core technique for remainder problems

✍️ MCQ

Choose one

Why does subtracting the remainder give us a number that

N

divides exactly?

✓

This works in general: if $N$ divides any number leaving remainder $r$ The starting condition - division with a remainder, then $N$ divides (number $- r$ ) exactlySubtract the remainder to get an exact multiple.

In our case: $N$ divides $245$ leaving remainder $5$ , so $N$ divides $245 - 5 = 240$ exactly.

✍️ MCQ

Choose one

If a number

N

divides

378

leaving remainder

8

, which number does

N

divide exactly?

✓

The subtraction converts an inexact division into an exact one.This is the core idea you need to remember

Why? The remainder $r$ is precisely the "extra bit" that prevents the original number from being a perfect multiple of $N$ Remove it and you're guaranteed a clean multiple. Remove it, and you're left with a clean multiple.

So the technique for remainder-adjusted problems:

Subtract the remainder from each given number.Step one: subtract first
Find HCF of the adjusted numbers.Step two: compute HCF

This two-step methodTwo simple steps make up the whole method is your go-to approachUse this whenever you see remainder problems whenever a problem asks for "the largest number that divides... leaving remainder..."

The adjusted numbers are the 'clean' multiplesNumbers N divides with zero remainder that $N$ divides exactly.

In our example: $245 - 5 = 240$ , and $N$ divides $240$ exactly with no remainder.

So instead of working with $245$ (which has that pesky remainder $5$ ), we work with $240$ — a clean multiple of $N$ .

✍️ MCQ

Choose one

A problem says: "Find the largest number that divides

378

and

294

, leaving remainder

6

in each case." What are the two adjusted numbers you should find the HCF of?

✓

2. Same remainder, known: full problem execution

You understand the principle. Now let's execute it on a complete problem — both numbers have the same known remainder, and you need to find the largest divisor.

Remember: If $N$ divides each number leaving remainder $r$ , then $N$ divides $(number - r)$ exactly for each.

So find HCF of the adjusted numbers.

✍️ Question

Problem:

Find the largest number that divides 245 and 1037, leaving remainder 5 in each case.

Show:

The subtraction step
The HCF computation
Verify your answer

Step 1 — Subtract the remainderSubtract remainder before doing HCF calculation from each number:

$245 - 5 =$ $240$ Numbers become exact multiples of N

$1037 - 5 =$ $1032$ Numbers become exact multiples of N

Now we have clean multiplesFinding HCF becomes straightforward to work with.

✍️ MCQ

Choose one

Why do we subtract 5 from both numbers before finding the HCF?

✓

Step 2 — Find HCF(240, 1032).

Let's start by factorising 240.

Factorise: $240 = 2^4 \times 3 \times 5$

For 1032, let's divide step by step:

$1032 \div 2 = 516$ Start with 2, the smallest prime

$516 \div 2 = 258$

$258 \div 2 = 129$

(prime!) $129 \div 3 = 43$ (129 is odd, so try the next prime)

So $1032 = 2^{3} \times 3 \times 43$

✍️ MCQ

Choose one

Which primes appear in both

240 = 2^4 \times 3 \times 5

and

1032 = 2^3 \times 3 \times 43

?

✓

Common primesOnly take primes found in both at lowest powersChoose the smaller exponent:

For 2: $\min (4, 3) = 3 \to$ $2^{3} = 8$
For 3: $\min (1, 1) = 1 \to 3$
5 only in 240, 43 only in 1032 — exclude bothPrimes in only one number are excluded

(Answer!)HCF = $8 \times 3 = 24$ (The greatest factor they share)

Step 3 — VerifyAlways check your answer:

$245 \div 24 = 10$ remainder $5$ Check you get remainder 5 (since $24 \times 10 = 240$ , and $245 - 240 = 5$ Subtract to find remainder). ✓ Correct.

$1037 \div 24 = 43$ remainder $5$ Both numbers leave remainder 5 (since $24 \times 43 = 1032$ , and $1037 - 1032 = 5$ Check the second number too). ✓ Correct.

Our answer 24 is verified!Now you know your answer is right

✍️ MCQ

Choose one

Why did we subtract

5

from both numbers before finding the HCF?

✓

⚠️ The common mistake: Computing HCF(245, 1037)This direct computation is the mistake directly, which gives HCF = 1wrong!Wrong answer because we skipped a step.

That's wrong — these numbers are coprimeNumbers with no common factor besides 1, but that's not what the question asks.

The question asks for the number that divides them with remainder 5Your signal to subtract before computing HCF.

✍️ MCQ

Choose one

If a question asks for the largest number that divides

100

and

250

leaving remainder

4

, what should you compute?

✓

3. Different remainders: subtract each one separately

Here's an interesting twist 🔄

Sometimes a problem gives you different remainders for different numbers.

For example:

A number $N$ might leave remainder 3 when dividing one number
But remainder 5 when dividing another

The principle is the same — subtract each remainder from its own number — but you need to be careful to apply the right subtraction to each numbercareful!.

📋 Given Info

Quick reminder:

If $N$ divides 129 leaving remainder 3, then $N$ divides $129 - 3 = 126$ exactly.
If $N$ divides 545 leaving remainder 5, then $N$ divides $545 - 5 = 540$ exactly.

Let's see if you can apply this!

✍️ Question

Problem 📝

Find the largest number that divides 129 and 545, leaving remainders 3r₁ and 5r₂ respectively.

Show your work.

When remainders are differentThis triggers the adjustment step, subtract EACH remainder from its OWN numberKey adjustment step before finding HCF.

$N$ divides 129 leaving remainder 3: $129 - 3 = 126$ . So $N$ divides 126clean exactly.
$N$ divides 545 leaving remainder 5: $545 - 5 = 540$ . So $N$ divides 540clean exactly.

✍️ MCQ

Choose one

Since

N

divides both 126 and 540 exactly, what should we find to get the largest possible

N

?

✓

This means $N$ is a common factor of 126 and 540. The largestThe largest common factor is your answer such $N$ = HCF(126, 540)Find HCF of the clean multiples.

Factorise:

$126 = 2 \times 3^2 \times 7$

$540 = 2^2 \times 3^3 \times 5$

✍️ MCQ

Choose one

Which prime factors are common to both

126

and

540

?

✓

Common primes at lowest powersTake the lowest power of each common prime:

2: $\min(1, 2) = 1$ minTake the minimum exponent
3: $\min(2, 3) = 2$ minTake the minimum exponent

HCF $= 2^1 \times 3^2$ (The largest number giving those remainders) $= 2 \times 9 =$ (answer) $18$ (Largest number that leaves those specific remainders)

✍️ FIB

Fill in the blank

What remainder do you get when you divide

129

by

18

?

✓

Verify:Always check your answer works

$129 \div 18 = 7$ remainder $3$ $(18 \times 7 = 126,$ $129 - 126 = 3)$ ✓verifiedCheck remainder matches what was asked

$545 \div 18 = 30$ remainder $5$ $(18 \times 30 = 540,$ $545 - 540 = 5)$ ✓verifiedConfirms answer gives required remainders

✍️ MCQ

Choose one

Why did we subtract different remainders from each number (

3

from

129

, and

5

from

545

) instead of subtracting the same number from both?

✓

The key:Each number has its own subtraction each number gets its OWN subtractionBased on its own remainder value. Don't subtract 3 from both, and don't subtract 5 from both.Common mistake students make

4. Unknown same remainder: the pairwise-difference trick

Here's the trickiest variant of remainder problems: the problem says the remainder is the same in each case but doesn't tell you what it is.

You can't subtract something you don't know.

But there's an elegant algebraic trick that eliminates the unknown remainder entirely.

📋 Given Info

Here's the setup:

If $N$ divides 43, 91, and 183 all leaving the same remainder $r$ , then:

$43 = N q_{1} + r$ $91 = N q_{2} + r$ $183 = N q_{3} + r$

Subtracting pairs:

$91 - 43 = N(q_2 - q_1)$
$183 - 91 = N(q_3 - q_2)$

The unknown $r$ cancels out!

✍️ Question

Your Turn 🧠

Find the greatest number that divides 43, 91, and 183 so as to leave the same remainder in each case.

In your answer:

Show the pairwise differences
Explain why they eliminate the unknown remainder
Find the HCF of the differences
Verify by checking the actual remainder when you divide each original number

When the remainder is unknownYou can't subtract it directly, you can't subtract it directly. But you CAN eliminate it by taking differencesSubtract to eliminate the unknown remainder.

If $N$ divides 43 leaving remainder $r$ : $43 = N q_{1} + r$

If $N$ divides 91 leaving remainder $r$ : $91 = N q_{2} + r$

$r$ keySame remainder makes the subtraction trick work

✍️ MCQ

Choose one

If we subtract the first equation from the second, what happens to the

r

terms?

✓

Subtract: $91 - 43 = N (q_{2} - q_{1})$

The $r$ terms cancelSubtraction eliminates the unknown remainder! So $N$ divides $48$ No unknown remainder anymore.

Similarly:

$183 - 91 =$ $92$ , and $N$ divides 92
$183 - 43 =$ $140$ , and $N$ divides 140

So $N$ must divide all three differencesWhen remainders cancel out, N divides the differences: 48, 92, and 140.

Therefore: $N = \text{HCF}(48, 92, 140)$ AnswerFind HCF of differences — that's the whole method

✍️ MCQ

Choose one

Why do we take the HCF of the differences (48, 92, 140) rather than the HCF of the original numbers (43, 91, 183)?

✓

FactoriseThe standard approach for finding HCF each difference:

$48 = 2^4 \times 3$ 48 broken into its prime factors
$92 = 2^2 \times 23$ (92 broken into its prime factors)
$140 = 2^2 \times 5 \times 7$ 140 broken into its prime factors

Find the common prime factorsMust be in every single number:

Common prime: only 2only one!(The only prime present in all three)

Minimum power across all three: $\min(4, 2, 2) = 2$ Pick the lowest exponent of the common prime

$∴ HCF = 2^{2} =$ $\boxed{4}$ The final HCF result

✍️ MCQ

Choose one

We found that only the prime

2

is common to all three factorisations. Why didn't

3

make it into the HCF?

✓

VerifyAlways check your HCF actually works:

$43 \div 4 = 10$ remainder $3$ $(4 \times 10 = 40, 43 - 40 =$ $3$ ✓Same remainder confirms correct answer $)$
$91 \div 4 = 22$ remainder $3$ $(4 \times 22 = 88, 91 - 88 =$ $3$ ✓Same remainder confirms correct answer $)$
$183 \div 4 = 45$ remainder $3$ $(4 \times 45 = 180, 183 - 180 =$ $3$ ✓Same remainder confirms correct answer $)$

All three numbers leave remainder 3 when divided by 4.

The unknown remainder turns out to be 3, and the answer is 4.

✨ Notice: We found the answer without ever knowing the remainder beforehand!We never needed to know the remainder first The subtraction trickSubtracting pairs makes the remainder cancel out eliminated it completely.

✍️ MCQ

Choose one

In this problem, why did we subtract pairs of numbers (like

91 - 43

) instead of subtracting a known remainder from each number?

✓

The technique summarised:

When the remainder is givenYour cue to subtract first, subtract it from each number firstDon't jump into HCF directly, then find the HCF.

Example: If $N$ divides 245 and 1037 each leaving remainder 5, compute $245 - 5 = 240$ HCF only works properly with multiples and $1037 - 5 = 1032$ HCF only works properly with multiples, then find $HCF (240, 1032)$ .

When the remainder is unknown but the sameThe trickier case for all numbers, take pairwise differencesThe remainder cancels out completely to cancel it, then find the HCF of those differences.

This is exactly what we did with 43, 91, and 183 — the differences $91 - 43 = 48$ , $183 - 91 = 92$ , and $183 - 43 = 140$ gave us $\text{HCF}(48, 92, 140) = 4$ HCFUnknown remainder means take differences first.

✍️ MCQ

Choose one

A problem says: "Find the greatest number that divides 56 and 98 leaving remainder 2 in each case." Which technique should you use?

✓