Find the HCF


TRICK to find HCF Quickly (Euclid Method)


Golden Trick: āϏ⧋āĻŖāĻžāϞ⧀ āĻ•ā§ŒāĻļāϞ:



  1. Keep dividing bigger number by smaller number. āĻĄāĻžāϙ⧰ āϏāĻ‚āĻ–ā§āϝāĻžāĻŸā§‹āĻ• āϏ⧰⧁ āϏāĻ‚āĻ–ā§āϝāĻžā§°ā§‡ āĻ­āĻžāĻ— āϕ⧰āĻ•

  2. When remainder becomes 0āϝ⧇āϤāĻŋāϝāĻŧāĻž āĻ…ā§ąāĻļāĻŋāĻˇā§āϟ 0 āĻšā§Ÿ,

  3. The last non-zero remainder = HCF. āϤāĻžā§° āφāĻ—ā§° āĻ…āĻļā§‚āĻ¨ā§āϝ āĻ…ā§ąāĻļāĻŋāĻˇā§āϟāĻŸā§‹ = HCF


Real Numbers


Q1. Find the HCF of 306 and 657 using Euclid’s Division Algorithm. āĻĒā§ā§°āĻļā§āύ: āχāωāĻ•ā§āϞāĻŋāĻĄā§° āĻ­āĻžāĻ— āĻāϞāĻ—ā§°āĻŋāĻĻāĻŽ āĻŦā§āĻ¯ā§ąāĻšāĻžā§° āϕ⧰āĻŋ 306 āφ⧰⧁ 657 ā§° HCF āωāϞāĻŋāϝāĻŧāĻžāĻ“āĻ•āĨ¤


Soln / āϏāĻŽāĻžāϧāĻžāύ


Using Euclid’s Division Algorithm,āχāωāĻ•ā§āϞāĻŋāĻĄā§° āĻ­āĻžāĻ— āĻāϞāĻ—ā§°āĻŋāĻĻāĻŽ āĻ…āύ⧁āϏ⧰āĻŋ -


657 = 306 × 2 + 45 , 1st Divide the bigger number by the smaller number, Here remainder 45.  āĻĄāĻžāϙ⧰ āϏāĻ‚āĻ–ā§āϝāĻžāĻŸā§‹āĻ• āϏ⧰⧁ āϏāĻ‚āĻ–ā§āϝāĻžā§°ā§‡ āĻ­āĻžāĻ— āϕ⧰āĻ•


306 = 45 × 6 + 36 , 2nd Now divide the previous divisor by the remainder, Here remainder 36. āĻāϤāĻŋāϝāĻŧāĻž āφāĻ—ā§° āĻ­āĻžāĻ—āĻ•ā§°ā§āϤāĻžāĻ• āĻ…ā§ąāĻļāĻŋāĻˇā§āĻŸā§°ā§‡ āĻ­āĻžāĻ— āϕ⧰āĻ•



36 = 9 × 4 + 0,   Continue till the remainder becomes 0. Here remainder 0. āĻ…ā§ąāĻļāĻŋāĻˇā§āϟ 0 āύāĻšā§‹ā§ąāĻžāϞ⧈āϕ⧇ āϚāϞāĻžāχ āϝāĻžāĻ“āĻ•


Since the remainder is 0, the last non-zero remainder is the HCF. āϝāĻŋāĻšā§‡āϤ⧁ āĻ…ā§ąāĻļāĻŋāĻˇā§āϟ 0, āϏ⧇āϝāĻŧ⧇ āĻļ⧇āώ⧰ āĻ…āĻļā§‚āĻ¨ā§āϝ āĻ…ā§ąāĻļāĻŋāĻˇā§āϟāĻŸā§‹ āĻšā§‡ HCFāĨ¤


HCF = 9, āĻ—.āϏāĻž.āϗ⧁ = 9


Q2. Find the HCF ofof 135 and 225 using Euclid’s Division Algorithm. āĻĒā§ā§°āĻļā§āύ 2: āχāωāĻ•ā§āϞāĻŋāĻĄā§° āĻ­āĻžāĻ— āĻāϞāĻ—ā§°āĻŋāĻĻāĻŽ āĻŦā§āĻ¯ā§ąāĻšāĻžā§° āϕ⧰āĻŋ 135 āφ⧰⧁ 225 ā§° HCF āωāϞāĻŋāϝāĻŧāĻžāĻ“āĻ•āĨ¤


Soln / āϏāĻŽāĻžāϧāĻžāύ



135 = 90 × 1 + 45, remainder 45


90 = 45 × 2 + 0, remainder 0


Last non-zero remainder = 45


HCF = 45, āĻ—.āϏāĻž.āϗ⧁ = 45


Q3. Find the HCF of 867 and 255 using Euclid’s Division Algorithm. āĻĒā§ā§°āĻļā§āύ 3: āχāωāĻ•ā§āϞāĻŋāĻĄā§° āĻ­āĻžāĻ— āĻāϞāĻ—ā§°āĻŋāĻĻāĻŽ āĻŦā§āĻ¯ā§ąāĻšāĻžā§° āϕ⧰āĻŋ 867 āφ⧰⧁ 255 ā§° HCF āωāϞāĻŋāϝāĻŧāĻžāĻ“āĻ•āĨ¤


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.āϗ⧁ = 51


Q4. Find the HCF of 420 and 156 using Euclid’s Division Algorithm. āĻĒā§ā§°āĻļā§āύ 4: āχāωāĻ•ā§āϞāĻŋāĻĄā§° āĻ­āĻžāĻ— āĻāϞāĻ—ā§°āĻŋāĻĻāĻŽ āĻŦā§āĻ¯ā§ąāĻšāĻžā§° āϕ⧰āĻŋ 420 āφ⧰⧁ 156 ā§° HCF āωāϞāĻŋāϝāĻŧāĻžāĻ“āĻ•āĨ¤


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.āϗ⧁ = 12


Q5. Find the HCF of 128 and 72 using Euclid’s Division Algorithm. āĻĒā§ā§°āĻļā§āύ 5: āχāωāĻ•ā§āϞāĻŋāĻĄā§° āĻ­āĻžāĻ— āĻāϞāĻ—ā§°āĻŋāĻĻāĻŽ āĻŦā§āĻ¯ā§ąāĻšāĻžā§° āϕ⧰āĻŋ 128 āφ⧰⧁ 72 ā§° HCF āωāϞāĻŋāϝāĻŧāĻžāĻ“āĻ•āĨ¤


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.āϗ⧁ = 8