Find the HCF
TRICK to find HCF Quickly (Euclid Method)
Golden Trick: āϏā§āĻŖāĻžāϞ⧠āĻā§āĻļāϞ:
- Keep dividing bigger number by smaller number. āĻĄāĻžāĻā§° āϏāĻāĻā§āϝāĻžāĻā§āĻ āϏ⧰⧠āϏāĻāĻā§āϝāĻžā§°ā§ āĻāĻžāĻ āĻā§°āĻ
- When remainder becomes 0, āϝā§āϤāĻŋāϝāĻŧāĻž āĻ ā§ąāĻļāĻŋāώā§āĻ 0 āĻšā§,
- The last non-zero remainder = HCF. āϤāĻžā§° āĻāĻā§° āĻ āĻļā§āύā§āϝ āĻ ā§ąāĻļāĻŋāώā§āĻāĻā§ = HCF
Real Numbers
Q1. Find the HCF of 306 and 657 using Euclid’s Division Algorithm. āĻĒā§ā§°āĻļā§āύ: āĻāĻāĻā§āϞāĻŋāĻĄā§° āĻāĻžāĻ āĻāϞāĻā§°āĻŋāĻĻāĻŽ āĻŦā§āĻ¯ā§ąāĻšāĻžā§° āĻā§°āĻŋ 306 āĻā§°ā§ 657 ā§° HCF āĻāϞāĻŋāϝāĻŧāĻžāĻāĻāĨ¤
Soln / āϏāĻŽāĻžāϧāĻžāύ
Using Euclid’s Division Algorithm,āĻāĻāĻā§āϞāĻŋāĻĄā§° āĻāĻžāĻ āĻāϞāĻā§°āĻŋāĻĻāĻŽ āĻ āύā§āϏ⧰āĻŋ -
657 = 306 × 2 + 45 , 1st Divide the bigger number by the smaller number, Here remainder 45. āĻĄāĻžāĻā§° āϏāĻāĻā§āϝāĻžāĻā§āĻ āϏ⧰⧠āϏāĻāĻā§āϝāĻžā§°ā§ āĻāĻžāĻ āĻā§°āĻ
306 = 45 × 6 + 36 , 2nd Now divide the previous divisor by the remainder, Here remainder 36. āĻāϤāĻŋāϝāĻŧāĻž āĻāĻā§° āĻāĻžāĻāĻā§°ā§āϤāĻžāĻ āĻ ā§ąāĻļāĻŋāώā§āĻā§°ā§ āĻāĻžāĻ āĻā§°āĻ
36 = 9 × 4 + 0, Continue till the remainder becomes 0. Here remainder 0. āĻ ā§ąāĻļāĻŋāώā§āĻ 0 āύāĻšā§ā§ąāĻžāϞā§āĻā§ āĻāϞāĻžāĻ āϝāĻžāĻāĻ
Since the remainder is 0, the last non-zero remainder is the HCF. āϝāĻŋāĻšā§āϤ⧠āĻ ā§ąāĻļāĻŋāώā§āĻ 0, āϏā§āϝāĻŧā§ āĻļā§āώ⧰ āĻ āĻļā§āύā§āϝ āĻ ā§ąāĻļāĻŋāώā§āĻāĻā§ āĻšā§ HCFāĨ¤
HCF = 9, āĻ.āϏāĻž.āĻā§ = 9
Q2. Find the HCF ofof 135 and 225 using Euclid’s Division Algorithm. āĻĒā§ā§°āĻļā§āύ 2: āĻāĻāĻā§āϞāĻŋāĻĄā§° āĻāĻžāĻ āĻāϞāĻā§°āĻŋāĻĻāĻŽ āĻŦā§āĻ¯ā§ąāĻšāĻžā§° āĻā§°āĻŋ 135 āĻā§°ā§ 225 ā§° HCF āĻāϞāĻŋāϝāĻŧāĻžāĻāĻāĨ¤
Soln / āϏāĻŽāĻžāϧāĻžāύ
135 = 90 × 1 + 45, remainder 45
90 = 45 × 2 + 0, remainder 0
Last non-zero remainder = 45
HCF = 45, āĻ.āϏāĻž.āĻā§ = 45
Q3. Find the HCF of 867 and 255 using Euclid’s Division Algorithm. āĻĒā§ā§°āĻļā§āύ 3: āĻāĻāĻā§āϞāĻŋāĻĄā§° āĻāĻžāĻ āĻāϞāĻā§°āĻŋāĻĻāĻŽ āĻŦā§āĻ¯ā§ąāĻšāĻžā§° āĻā§°āĻŋ 867 āĻā§°ā§ 255 ā§° HCF āĻāϞāĻŋāϝāĻŧāĻžāĻāĻāĨ¤
Soln / āϏāĻŽāĻžāϧāĻžāύ
867 = 255 × 3 + 102, remainder 102
255 = 102 × 2 + 51, remainder 51
102 = 51 × 2 + 0, remainder 0
Last non-zero remainder = 51
HCF = 51, āĻ.āϏāĻž.āĻā§ = 51
Q4. Find the HCF of 420 and 156 using Euclid’s Division Algorithm. āĻĒā§ā§°āĻļā§āύ 4: āĻāĻāĻā§āϞāĻŋāĻĄā§° āĻāĻžāĻ āĻāϞāĻā§°āĻŋāĻĻāĻŽ āĻŦā§āĻ¯ā§ąāĻšāĻžā§° āĻā§°āĻŋ 420 āĻā§°ā§ 156 ā§° HCF āĻāϞāĻŋāϝāĻŧāĻžāĻāĻāĨ¤
Soln / āϏāĻŽāĻžāϧāĻžāύ
420 = 156 × 2 + 108 , remainder 108
156 = 108 × 1 + 48 , remainder 48
48 = 12 × 4 + 0 , remainder 0
Last non-zero remainder = 12
HCF = 12, āĻ.āϏāĻž.āĻā§ = 12
Q5. Find the HCF of 128 and 72 using Euclid’s Division Algorithm. āĻĒā§ā§°āĻļā§āύ 5: āĻāĻāĻā§āϞāĻŋāĻĄā§° āĻāĻžāĻ āĻāϞāĻā§°āĻŋāĻĻāĻŽ āĻŦā§āĻ¯ā§ąāĻšāĻžā§° āĻā§°āĻŋ 128 āĻā§°ā§ 72 ā§° HCF āĻāϞāĻŋāϝāĻŧāĻžāĻāĻāĨ¤
Soln / āϏāĻŽāĻžāϧāĻžāύ
128 = 72 × 1 + 56 , remainder 56
72 = 56 × 1 + 16 , remainder 16
56 = 16 × 3 + 8 , remainder 8
16 = 8 × 2 + 0 , remainder 0
Last non-zero remainder = 8
HCF = 8, āĻ.āϏāĻž.āĻā§ = 8