Real numbers - Euclid’s Division Algorithm for class 10th/grade X maths 2014. Lets tute is an online learning centre. We provide quality education for all learners and 24/7 academic guidance through E-tutoring. Our Mission- Our aspiration is to be a renowned unpaid school on Web-World

1. Problems based on

2. Greatest Common Divisor / Highest Common Factor (HCF)
Example: HCF for 2 & 8 = 2
• For two positive integers a & b where a > b , they can be expressed as
• Where 0 ≤ r < b & q € Ẕ .If “r = 0” then “b” is the HCF/GCD of “a & b”
• If “r ≠ 0” then apply Euclid’sdivision lemma to b and r.
• For some integers m and n , 0 ≤ n < r
• Continue this process till the remainder is zero.
• The divisor which gives you the remainder as 0 is your HCF/GCD
rqba  )(
nmrb  )(

3. 1. Find the HCF of 256 and 16 using Euclid’s Division Algorithm
• Let a = 256 and b = 16
When represented in form
We get: 256 = (16×16) + 0
r = 0
16 is the HCF of 256 & 16
Examples
16
0
256
25616
rqba  )(

4. 2. Find the HCF of 13 and 24 using Euclid’s Division Algorithm
Let, a = 24 and b = 13
“r ≠ 0” Again “r ≠ 0”
New dividend and divisor are  New dividend and divisor are 2 and 1
13 and 11 respectively
Again, “r ≠ 0” r = 0,  HCF of 13 & 24 is 1
New dividend and divisor are
11 and 2 respectively
1
11
13
2413
5
1
10
112
1
2
11
1311
2
0
2
21

5. 3. Find the HCF of 1424 and 3084 using Euclid’s Division Algorithm
Let, a = 3084 and b = 1424 .
“r ≠ 0” Again “r ≠ 0”
New dividend and divisor are  New dividend and divisor are 8 and 4.
1424 and 236 respectively.
Again, “r ≠ 0” r = 0,  HCF of 1424 & 3084 is 4.
New dividend and divisor are
236 and 8 respectively.
6
8
1416
1424236
2
236
2848
30841424
29
4
232
2368
2
0
8
84

6. 4. Find the largest number that divides 2623 & 2011 and leaves
Remainders 5 & 9 respectively
Point to remember
2 divides 25 leaving a remainder =1
Also, 2 divides 24 completely leaving a remainder = 0
24 = 25 – 1
Same logic has to be applied for this question
Q 2623 & 2011 when divided leaves remainders of 5 & 9
We have to find HFC of 2623 – 5 = 2618 and 2011 – 9 = 2002,
so we consider the numbers 2618 & 2002.
2618 = 2002 × 1 + 616
Now applying Euclid’s lemma to 2618 & 2002 we get,
As r ≠ 0 we again apply Euclid’s lemma to 2002 & 616.

7. 616 = 154 × 4 + 0
Now, Remainder (r) = 0
Hence, the required number is 154
We have 2002 = 616 × 3 + 154 as again we see that r ≠ 0.
Applying Euclid’s lemma again to 616 and 154 we get,
Hence, according to the algorithm the divisor = HCF/GCD
∴ 154 = HCF of 2618 and 2002

8. 5. A milkman has 175 liters of cow’s milk & 105 liters of goat’s milk .He
wants to sell the milk by filling the two types of milk in cans of equal
capacity. What should be the maximum capacity of such a can and how
many cans of cow’s milk and goat’s milk will be obtained after the
distribution?
175 liters
105 liters

9. Applying Euclid’s Division Algorithm for 175 and 105 we get:
175 = 105 × 1 + 70
The two types of milk have to be filled in cans having the same and
maximum possible capacity
The required capacity of such a can will be :
HCF of 175 and 105 (in liters)
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
1
70
105
175105
Remainder (r) ≠ 0
New divisor = 70 and Dividend = 105
Again applying Euclid’s Division Algorithm we get:
105 = 70 × 1 + 35
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
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1
35
70
10570

10. 70 = 35 × 2 + 0
Hence the required capacity of the can is 35 liters
Again, Remainder (r) ≠ 0
New divisor = 35 and Dividend = 70
Again applying Euclid’s Division Algorithm we get:

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



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
2
0
70
7035
Now, Remainder (r) = 0
Hence, according to the algorithm the divisor = HCF
 35 = HCF of 175 and 105
And after the distribution we will obtain 175/35 = 5 cans of cow’s milk
and 107/35 = 3 cans of goat’s milk.

11. REVISION
1. The divisor which gives us remainder (r) = 0 after division is the
HCF
2. Look for terms in the problem like “maximum” or “largest” which
indicates that you have to find the HCF
3. If a/b gives a remainder (r) ≠ 0 then; (a – r) is a number which is
completely divisible by b