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00:02
28 Mar 2026

Executing Euclid's Division Algorithm on Different Number Pairs

You now understand WHY Euclid's Division Algorithm works. The next challenge is executing it smoothly and accurately.

EDA is a procedure, and procedures require practice.

Each step involves a division of multi-digit numbers, and one arithmetic error in any step corrupts every step that follows.

In this section, you will:

  • Work through examples of increasing difficulty
  • Develop strategies for quick estimation of quotients

You will also encounter two important special cases:

  1. When the algorithm terminates in a single step
  2. When you accidentally start by dividing the smaller number by the larger

By the end, you should be able to execute EDA on any pair of numbers the CBSE board exam can throw at you, with confidence in both your speed and your accuracy.

TargetGoal
SpeedQuick estimation
AccuracyZero arithmetic errors

1. Full EDA execution — a three-step chain

Building EDA Fluency 🔢

The best way to build fluency with Euclid's Division Algorithm (EDA) is to execute it from start to finish on a fresh problem.

  • Find the HCF of the given pair of numbers.
  • Write every step as a numbered EDL equation (a=bq+ra = bq + r).

This chain will have three steps, which is typical of CBSE board exam questions.

✍️ Question

Your Turn ✏️

Execute Euclid's Division Algorithm (EDA) on the pair:

(1651, 2032)

  1. Number your steps clearly.
  2. Write each step as an EDL equation (a=bq+ra = bq + r).
  3. State the HCF clearly at the end.

Let me work through this step by step.

We need to find the Highest Common Factor, or HCF(1651,2032)HCF(1651, 2032).

✍️ MCQ
Choose one
In the lemma a=bq+ra = bq + r, what is the standard rule for choosing 'aa'?

Since 2032>16512032 > 1651, we divide 20322032 by 16511651.

✍️ MCQ
Choose one
Estimate the quotient for 2032÷16512032 \div 1651.

Step (i): 2032 ÷ 1651

How many times does 1651 fit into 2032?

  • 1651×1=16511651 \times 1 = 1651 (fits, since 1651<20321651 < 2032)
  • 1651×2=33021651 \times 2 = 3302 (too big, since 3302>20323302 > 2032)
✍️ FIB
Fill in the blank
What is the remainder for 203216512032 - 1651?
Type your answer, or hold Space to speak

So quotient = 1. Remainder = 20321651=3812032 - 1651 = 381.

Equation: 2032=1651×1+3812032 = 1651 \times 1 + 381

Verify: 1651+381=20321651 + 381 = 2032. ✓ Correct.

✍️ MCQ
Choose one
Which numbers become the new pair for the next step?

Remainder 381 is not 0, so continue. New pair: (1651, 381).

Step (ii): 1651÷3811651 \div 381

Following the algorithm, our previous divisor (1651) now becomes the dividend, and our previous remainder (381) becomes the new divisor.

✍️ MCQ
Choose one
Estimate the quotient: How many times does 381381 fit into 16511651?

How many times does 381 fit into 1651?

  • 381×4=1524381 \times 4 = 1524 (fits, since 1524<16511524 < 1651)
  • 381×5=1905381 \times 5 = 1905 (too big, since 1905>16511905 > 1651)

Mental Math Tip: Break down 381×4381 \times 4. 380×4=1520380 \times 4 = 1520 1×4=41 \times 4 = 4 Total = 1524

✍️ FIB
Fill in the blank
What is the remainder: 165115241651 - 1524?
Type your answer, or hold Space to speak

So, Quotient = 4 and Remainder = 127.

Equation: 1651=381×4+1271651 = 381 \times 4 + 127

Verify: 1524+127=16511524 + 127 = 1651. ✅ Correct.

Since the remainder 127 is not 0, we must continue.

New pair for Step (iii): (381, 127)

Step (iii): 381÷127381 \div 127

Earlier, we found our new divisor is 381381 and our remainder is 127127. Now, we see how many times 127127 fits into 381381.

✍️ MCQ
Choose one
Estimate the quotient for 381÷127381 \div 127:

127×3=381127 \times 3 = 381. Exactly!

So, quotient = 3 and remainder = 0.

Equation: 381=127×3+0381 = 127 \times 3 + 0

Since the remainder is 0, we STOP here.

✍️ MCQ
Choose one
The HCF is the divisor at the stage where the remainder becomes zero. What is the HCF here?

The Final Result

HCF(1651, 2032) = 127

As we just saw, 127 was the last divisor in the step where the remainder finally reached zero.

Verification

It is always good practice to verify our result:

  • 1651÷127=131651 \div 127 = 13 (integer)
  • 2032÷127=162032 \div 127 = 16 (integer)

Both divide evenly. ✓ Confirmed.

✍️ Yes/No
Yes or No?
Does the fact that both original numbers divide evenly by 127127 confirm that 127127 is a common factor?
✍️ MCQ
Choose one
If we had found that 1651÷1271651 \div 127 resulted in a decimal like 13.513.5, what would that tell us?

2. One-step termination — when one number divides the other

Not every EDA chain requires multiple steps. Sometimes the smaller number divides the larger one exactly, and the algorithm terminates immediately.

This is an important case to recognise because:

  • It frequently appears on exams
  • It reinforces the stopping condition (remainder = 00)

In such cases, the smaller of the two numbers is itself the HCF.

✍️ Question

Find HCF(196, 38220) using Euclid's Division Algorithm (EDA).

  • Execute EDA on the pair (196,38220)(196, 38220).
  • How many steps does it take?
  • What is the HCF?

A Surprising Single-Step Case

This is a special case that catches students off guard — they expect the algorithm to always require multiple steps.

Since 38220>19638220 > 196, we divide 3822038220 by 196196.

✍️ MCQ
Choose one
If the remainder is zero in the first division, how many steps does the algorithm take?

Estimation Strategy

To find the quotient, let us estimate:

  • 196196 is close to 200200.
  • 200×190=38000200 \times 190 = 38000, which is close to 3822038220.

So the quotient is around 195195.

✍️ MCQ
Choose one
Is 196×190196 \times 190 greater than or less than 3800038000?

Verifying the Calculation

Let me verify: 196×195196 \times 195.

Instead of long multiplication, let's break it down:

  • 196×200=39200196 \times 200 = 39200
  • 196×5=980196 \times 5 = 980
✍️ FIB
Fill in the blank
What is 3920098039200 - 980?
Type your answer, or hold Space to speak

39200980=3822039200 - 980 = 38220

So, 196×195=38220196 \times 195 = 38220 exactly.

The remainder is 00.

Here is our result in the form of the Euclid's Division Lemma (EDL) equation:

38220=196×195+038220 = 196 \times 195 + 0

Since the remainder is 00 right away, the algorithm stops here. The HCF is the divisor: 196196.

HCF(196,38220)=196\text{HCF}(196, 38220) = 196

✍️ MCQ
Choose one
In the equation a=bq+ra = bq + r, if r=0r = 0, what is HCF(a,b)\text{HCF}(a, b)?

This makes sense: 196196 divides 3822038220 exactly (38220÷196=19538220 \div 196 = 195), so 196196 is a factor of 3822038220. And the HCF of any number with one of its factors is that factor itself.

✍️ FIB
Fill in the blank
What is HCF(15,45)\text{HCF}(15, 45)?
Type your answer, or hold Space to speak

Exam Tip: Single-Step Solutions

On the exam, do not be surprised if the algorithm terminates in one step.

Just write the single equation, note that the remainder is 00, and state the HCF. You are done.

3. Starting order — what happens if you divide smaller by larger?

Quick scenario to think about 🤔

The textbook says to start Euclid's Division Algorithm with a>ba > b — divide the larger number by the smaller one.

But what if you accidentally mix up the order? Let's explore what happens — this reveals a nice self-correcting property of the algorithm.

📋 Given Info

Here's the situation:

A student wants to find HCF(45,120)\text{HCF}(45, 120).

  • Expected Step: Divide 120120 by 4545 (larger ÷\div smaller).
  • Accidental Step: They divide 4545 by 120120 (smaller ÷\div larger).

Will the algorithm still work, or will it break down?

✍️ Question

Your turn to investigate ✏️

a) Write the Euclid's Division Lemma equation for 4545 divided by 120120.

  • What are the quotient (qq) and remainder (rr)?

b) Based on your result, what pair of numbers would the student use for the next step of the algorithm?

c) Is the algorithm now on the correct track? What is the lesson here?

What happens when you divide a smaller number by a larger number?

Let us try a specific example to see how the math behaves: divide 45 by 120.

✍️ MCQ
Choose one
How many times does 120120 fit into 4545?

Since 120 is bigger than 45, it fits Zero times.

So, the quotient is 0.

To find the remainder: Remainder = 45(120×0)45 - (120 \times 0) Remainder = 450=4545 - 0 = 45

✍️ FIB
Fill in the blank
In the equation a=bq+ra = bq + r, if a=45a = 45 and b=120b = 120, what is the value of rr?
Type your answer, or hold Space to speak

Putting it all together into the EDL equation:

45=120×0+4545 = 120 \times 0 + 45

Notice that in the next step of the algorithm, the numbers will effectively 'swap' places.

Checking Validity

Let us check: is this a valid EDL equation?

The constraint says 0r<b0 \leq r < b.

✍️ Yes/No
Yes or No?
Is the remainder r=45r = 45 valid for the divisor b=120b = 120?

i.e., 045<1200 \leq 45 < 120. Yes, this is valid.

So this is a perfectly legal step — it just does not make progress toward finding the HCF.

✍️ MCQ
Choose one
What is the next division step?

Now, for the next step: our old divisor (120120) becomes the new dividend, and our old remainder (4545) becomes the new divisor.

This gives us the next pair: (120, 45).

✍️ MCQ
Choose one
What is the next division step?

But notice something—(120,45)(120, 45) is the pair we should have started with in the first place!

So the student's first step was effectively 'wasted'—it didn't perform any actual division, it just swapped the numbers into the correct order (a>ba > b) for us.

✍️ Yes/No
Yes or No?
Does starting with the smaller number first change the final HCF result?

As we discussed, our first step effectively swapped the numbers. Now, we continue correctly with the larger number as the dividend:

120=45×2+30120 = 45 \times 2 + 30

✍️ MCQ
Choose one
Which pair will be used in the next division step?

Exactly! We move forward with 4545 and 3030, then keep going until the remainder is zero:

45=30×1+1545 = 30 \times 1 + 15

30=15×2+030 = 15 \times 2 + 0

HCF(45, 120) = 15.

✍️ FIB
Fill in the blank
The HCF of 4545 and 120120 is ___
Type your answer, or hold Space to speak

The algorithm produced the correct answer — it just took 4 steps instead of 3.

Key Takeaway: The self-correcting property is nice to know (you will never get a wrong answer from a wrong starting order), but starting with the larger number is better because it saves a step.

4. Verifying each EDA step to catch arithmetic errors

Catching Errors in EDA 🔍

Arithmetic errors are the most common source of wrong answers in Euclid's Division Algorithm problems.

  • Since each step feeds into the next, one error cascades through the entire chain.
  • A mistake in the first remainder will make every subsequent step incorrect.

The Antidote: Verify each equation immediately after writing it.

📋 Given Info

A student is finding HCF(4052, 420) and writes:

  • Step 1: 4052=420×9+2724052 = 420 \times 9 + 272
  • Step 2: 420=272×1+158420 = 272 \times 1 + 158
  • Step 3: 272=158×1+114272 = 158 \times 1 + 114

(continues from here...)

Let's check if these steps are correct before proceeding.

✍️ Question

Your task: 🧮

(a) Verify Step 1 by computing 420×9+272420 \times 9 + 272. Is it correct?

(b) Verify Step 2 by computing 272×1+158272 \times 1 + 158. Is it correct?

(c) If any step is wrong, find the correct equation for that step.

Let me show you the verification process for each step.

The verification rule:

For any equation a=b×q+ra = b \times q + r, check that b×q+rb \times q + r actually equals aa.

✍️ Yes/No
Yes or No?
Does 9×5+2=479 \times 5 + 2 = 47?

Why verify?

Because the HCF is the last non-zero divisor, one wrong remainder in step 1 will lead to the wrong HCF in the final step.

Remember our earlier example: 1651=381×4+1271651 = 381 \times 4 + 127

If we check: 381×4+127=1524+127=1651381 \times 4 + 127 = 1524 + 127 = 1651. Matches! ✓

Step 1: 4052=420×9+2724052 = 420 \times 9 + 272

✍️ MCQ
Choose one
What is 420×9420 \times 9?

Compute 420×9420 \times 9:

  • 400×9=3600400 \times 9 = 3600
  • 20×9=18020 \times 9 = 180
  • Total =3780= 3780
✍️ Yes/No
Yes or No?
Does 3780+2723780 + 272 equal 40524052?

Now add the remainder: 3780+272=40523780 + 272 = 4052.

Does this equal the dividend? 4052=40524052 = 4052. YES. Step 1 is correct.

Step 2: Verification

Step 2: 420=272×1+158420 = 272 \times 1 + 158

✍️ FIB
Fill in the blank
Calculate 272+158272 + 158.
Type your answer, or hold Space to speak

Compute 272×1272 \times 1: 272272.

Now add the remainder: 272+158=430272 + 158 = 430.

✍️ Yes/No
Yes or No?
Does 430430 equal the dividend 420420?

Does this equal the dividend? 430430 vs 420420.

NO! 430420430 \neq 420.

Step 2 is WRONG.

The error is in the remainder. The correct computation is:

r=420(272×1)=420272=148r = 420 - (272 \times 1) = 420 - 272 = 148

✍️ MCQ
Choose one
Which is the correct equation for Step 2?

So the correct Step 2 is:

420=272×1+148420 = 272 \times 1 + 148

Verify: 272+148=420272 + 148 = 420. ✅ Correct.

The student wrote 158158 instead of 148148 — a simple subtraction error (420272=148420 - 272 = 148, not 158158).

The Danger: This one error would corrupt every subsequent step and eventually give the wrong HCF.

The 10-Second Rule

After writing each EDA equation, take 10 seconds to verify it.

The Verification Step:

  1. Multiply the divisor by the quotient.
  2. Add the remainder.
  3. Check if the total equals the dividend.
✍️ MCQ
Choose one
To verify the step 420=272×1+148420 = 272 \times 1 + 148, which calculation would you perform?

This quick check catches errors before they cascade.

As we saw in the previous example, one small arithmetic mistake in the remainder can ruin the entire process. Verification keeps your solution on the right track.

Exam Strategy

On the exam, you do not need to write out the verification — just do it mentally or in the margin.

Important: Do it for every step, not just the hard ones!

✍️ Yes/No
Yes or No?
Is it mandatory to show the verification calculations as part of your final answer in the exam?

5. Complete EDA on a fresh problem with verification

Time to put everything together!

You have learned why Euclid's Division Algorithm works and how to verify your steps. Now, let's see if you can execute the full algorithm on a problem you haven't seen before.

This is exactly what you would face on a board exam:

  1. Execute the full algorithm step-by-step.
  2. Verify each equation using the "10-second check."
  3. State the final HCF clearly.
✍️ Question

Find HCF(306, 657) using Euclid's Division Algorithm

Execute the full EDA.

  1. Write each step as a numbered EDL equation (a=bq+ra = bq + r).
  2. Show your work clearly for each division.
  3. State the final HCF once the remainder reaches zero.

Starting the HCF Journey

Let me walk through this process carefully, verifying every step as we go.

Our goal is to find HCF(306, 657).

✍️ MCQ
Choose one
Which number should be the dividend (aa)?

That's right! Since 657>306657 > 306, we follow the rule and divide the larger number by the smaller one.

So, our first step is to divide 657657 by 306306.

✍️ MCQ
Choose one
Estimate the quotient: 657÷306657 \div 306

(i) 657 ÷ 306:

306×2=612306 \times 2 = 612 (fits, since 612<657612 < 657)

306×3=918306 \times 3 = 918 (too big, since 918>657918 > 657)

✍️ FIB
Fill in the blank
Calculate the remainder: 657612657 - 612
Type your answer, or hold Space to speak

Quotient = 2. Remainder = 657612=45657 - 612 = 45.

Equation: 657=306×2+45657 = 306 \times 2 + 45.

Verify: 612+45=657612 + 45 = 657. ✓ Correct.

✍️ MCQ
Choose one
What is the next pair for the EDA?

Remainder 45 is not 0. Next pair: (306, 45).

(ii) 306 ÷ 45

Let's test our estimates for the division:

  • 45×6=27045 \times 6 = 270 (fits, since 270<306270 < 306)
  • 45×7=31545 \times 7 = 315 (too big, since 315>306315 > 306)
✍️ MCQ
Choose one
What is the quotient for 306÷45306 \div 45?

Quotient = 6.

Remainder = 306270=36306 - 270 = 36.

The Equation: 306=45×6+36306 = 45 \times 6 + 36

✍️ Yes/No
Yes or No?
Does 45×6+3645 \times 6 + 36 equal 306306?

Verify: 270+36=306270 + 36 = 306. ✓ Correct.

Since the remainder 36 is not 0, we must continue.

Next pair: (45, 36).

Step (iii): 45÷3645 \div 36

As we just decided, our new pair is 4545 and 3636. Let's find out how many times 3636 goes into 4545.

  • 36×1=3636 \times 1 = 36 (fits, since 36<4536 < 45)
  • 36×2=7236 \times 2 = 72 (too big, since 72>4572 > 45)
✍️ FIB
Fill in the blank
What is the remainder for 45÷3645 \div 36?
Type your answer, or hold Space to speak

Quotient = 1. Remainder = 4536=945 - 36 = 9.

The Equation: 45=36×1+945 = 36 \times 1 + 9

Verify: 36+9=4536 + 9 = 45. ✓ Correct.

✍️ MCQ
Choose one
Identify the next pair for the EDA:

Remainder 9 is not 0.

Next pair: (36, 9).

Step (iv): The Final Division

(iv) 36 ÷ 9:

9×4=369 \times 4 = 36. Exactly!

✍️ MCQ
Choose one
If 9×4=369 \times 4 = 36, what is the remainder?

Quotient = 4. Remainder = 0.

Equation: 36=9×4+036 = 9 \times 4 + 0

Verify: 36+0=3636 + 0 = 36. ✓ Correct.

✍️ MCQ
Choose one
What is the next step when the remainder is 00?

Remainder is 0. STOP.

HCF(306, 657) = 9.

Final verification: Does 9 divide both original numbers?

✍️ FIB
Fill in the blank
What is 306÷9306 \div 9?
Type your answer, or hold Space to speak

306÷9=34306 \div 9 = 34 (integer). ✓ Yes.

✍️ MCQ
Choose one
What is 657÷9657 \div 9?

657÷9=73657 \div 9 = 73 (integer). ✓ Yes.

Confirmed: 9 is the HCF.