Real numbers - Euclid’s Division Algorithm for class 10th/grade X maths 2014

REAL NUMBERS

Problems based on

EUCLID’S DIVISION ALGORITHM

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

Euclid’s Division Algorithm

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REAL NUMBERSwww.letstute.co

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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

025625616

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2. Find the HCF of 13 and 24 using Euclid’s Division AlgorithmLet, a = 24 and b = 13

“r ≠ 0” Again “r ≠ 0”New dividend and divisor are New dividend and divisor are 2 and 113 and 11 respectively

Again, “r ≠ 0” r = 0, HCF of 13 & 24 is 1\New dividend and divisor are 11 and 2 respectively

1

11132413

5

110112

1

2111311

2

0221



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3. Find the HCF of 1424 and 3084 using Euclid’s Division AlgorithmLet, 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

814161424236

2

236284830841424

29

42322368

2

0884



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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 = 024 = 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.



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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.

616 = 154 × 4 + 0

Now, Remainder (r) = 0

Hence, the required number is 154



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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

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



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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)

1

70105175105

Remainder (r) ≠ 0

New divisor = 70 and Dividend = 105

Again applying Euclid’s Division Algorithm we get:

105 = 70 × 1 + 35

1

3570

10570



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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:

2

0707035

Now, Remainder (r) = 0

Hence, according to the algorithm the divisor = HCF

35 = HCF of 175 and 105



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And after the distribution we will obtain 175/35 = 5 cans of cow’s milk and 107/35 = 3 cans of goat’s milk.

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



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Real numbers - Euclid’s Division Algorithm for class 10th/grade X maths 2014

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