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:
You will also encounter two important special cases:
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.
| Target | Goal |
|---|---|
| Speed | Quick estimation |
| Accuracy | Zero arithmetic errors |
The best way to build fluency with Euclid's Division Algorithm (EDA) is to execute it from start to finish on a fresh problem.
This chain will have three steps, which is typical of CBSE board exam questions.
Execute Euclid's Division Algorithm (EDA) on the pair:
(1651, 2032)
Let me work through this step by step.
We need to find the Highest Common Factor, or .
Since , we divide by .
Step (i): 2032 ÷ 1651
How many times does 1651 fit into 2032?
So quotient = 1. Remainder = .
Equation:
Verify: . ✓ Correct.
Remainder 381 is not 0, so continue. New pair: (1651, 381).
Following the algorithm, our previous divisor (1651) now becomes the dividend, and our previous remainder (381) becomes the new divisor.
How many times does 381 fit into 1651?
Mental Math Tip: Break down . Total = 1524
So, Quotient = 4 and Remainder = 127.
Equation:
Verify: . ✅ Correct.
Since the remainder 127 is not 0, we must continue.
New pair for Step (iii): (381, 127)
Earlier, we found our new divisor is and our remainder is . Now, we see how many times fits into .
. Exactly!
So, quotient = 3 and remainder = 0.
Equation:
Since the remainder is 0, we STOP here.
HCF(1651, 2032) = 127
As we just saw, 127 was the last divisor in the step where the remainder finally reached zero.
It is always good practice to verify our result:
Both divide evenly. ✓ Confirmed.
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:
In such cases, the smaller of the two numbers is itself the HCF.
Find HCF(196, 38220) using Euclid's Division Algorithm (EDA).
This is a special case that catches students off guard — they expect the algorithm to always require multiple steps.
Since , we divide by .
To find the quotient, let us estimate:
So the quotient is around .
Let me verify: .
Instead of long multiplication, let's break it down:
So, exactly.
The remainder is .
Here is our result in the form of the Euclid's Division Lemma (EDL) equation:
Since the remainder is right away, the algorithm stops here. The HCF is the divisor: .
This makes sense: divides exactly (), so is a factor of . And the HCF of any number with one of its factors is that factor itself.
On the exam, do not be surprised if the algorithm terminates in one step.
Just write the single equation, note that the remainder is , and state the HCF. You are done.
The textbook says to start Euclid's Division Algorithm with — 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.
A student wants to find .
Will the algorithm still work, or will it break down?
a) Write the Euclid's Division Lemma equation for divided by .
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?
Let us try a specific example to see how the math behaves: divide 45 by 120.
Since 120 is bigger than 45, it fits Zero times.
So, the quotient is 0.
To find the remainder: Remainder = Remainder =
Putting it all together into the EDL equation:
Notice that in the next step of the algorithm, the numbers will effectively 'swap' places.
Let us check: is this a valid EDL equation?
The constraint says .
i.e., . Yes, this is valid.
So this is a perfectly legal step — it just does not make progress toward finding the HCF.
Now, for the next step: our old divisor () becomes the new dividend, and our old remainder () becomes the new divisor.
This gives us the next pair: (120, 45).
But notice something— 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 () for us.
As we discussed, our first step effectively swapped the numbers. Now, we continue correctly with the larger number as the dividend:
Exactly! We move forward with and , then keep going until the remainder is zero:
HCF(45, 120) = 15.
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.
Arithmetic errors are the most common source of wrong answers in Euclid's Division Algorithm problems.
The Antidote: Verify each equation immediately after writing it.
A student is finding HCF(4052, 420) and writes:
(continues from here...)
Let's check if these steps are correct before proceeding.
(a) Verify Step 1 by computing . Is it correct?
(b) Verify Step 2 by computing . 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.
For any equation , check that actually equals .
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:
If we check: . Matches! ✓
Step 1:
Compute :
Now add the remainder: .
Does this equal the dividend? . YES. Step 1 is correct. ✓
Step 2:
Compute : .
Now add the remainder: .
Does this equal the dividend? vs .
NO! .
Step 2 is WRONG. ✗
The error is in the remainder. The correct computation is:
So the correct Step 2 is:
Verify: . ✅ Correct.
The student wrote instead of — a simple subtraction error (, not ).
The Danger: This one error would corrupt every subsequent step and eventually give the wrong HCF.
After writing each EDA equation, take 10 seconds to verify it.
The Verification Step:
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.
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!
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:
Execute the full EDA.
Let me walk through this process carefully, verifying every step as we go.
Our goal is to find HCF(306, 657).
That's right! Since , we follow the rule and divide the larger number by the smaller one.
So, our first step is to divide by .
(i) 657 ÷ 306:
(fits, since )
(too big, since )
Quotient = 2. Remainder = .
Equation: .
Verify: . ✓ Correct.
Remainder 45 is not 0. Next pair: (306, 45).
Let's test our estimates for the division:
Quotient = 6.
Remainder = .
The Equation:
Verify: . ✓ Correct.
Since the remainder 36 is not 0, we must continue.
Next pair: (45, 36).
As we just decided, our new pair is and . Let's find out how many times goes into .
Quotient = 1. Remainder = .
The Equation:
Verify: . ✓ Correct.
Remainder 9 is not 0.
Next pair: (36, 9).
(iv) 36 ÷ 9:
. Exactly!
Quotient = 4. Remainder = 0.
Equation:
Verify: . ✓ Correct.
Remainder is 0. STOP.
HCF(306, 657) = 9.
(integer). ✓ Yes.
(integer). ✓ Yes.
Confirmed: 9 is the HCF.