i. 45

∵ here 405 < 2520,

∴ b = 405 and a = 2520.

By Euclid’s division lemma -

a = bq + r ….(1)

where q is the quotient, r is the remainder and b is the divisor.

Putting it in equation (1) -

⇒ 2520 = 405(6) + 90.

Here 90 is the remainder, which is not zero. Again applying the Euclid’s divison lemma -

Now, a = 405; b = 90;

⇒ 405 = 90(4) + 45.

Here 45 is the remainder, which is not zero. Again applying the Euclid’s divison lemma -

a = 90; b = 45;

⇒ 90 = 45(2) + 0.

∵ remainder is zero.

∴ HCF is 45.

ii. 36

∵ here 504 < 1188,

∴ b = 504 and a = 1188.

By Euclid’s division lemma -

a = bq + r ….(1)

where q is the quotient, r is the remainder and b is the divisor.

Putting the values in equation (1) -

⇒ 1188 = 504(2) + 180.

Here 180 is the remainder, which is not zero. Again applying the Euclid’s divison lemma -

Now, a = 504; b = 180;

⇒ 504 = 180(2) + 144

Here 144 is the remainder, which is not zero. Again applying the Euclid’s divison lemma -

Now,a = 180; b = 144;

⇒ 180 = 144(1) + 36.

Here 36 is the remainder, which is not zero. Again applying the Euclid’s divison lemma -

Now,a = 144; b = 36;

⇒ 144 = 36(4) + 0.

∵ remainder is zero.

∴ HCF is 36.

iii. 15

∵ here 960 < 1575,

∴ b = 960 and a = 1575.

a = bq + r ….(1)

where q is the quotient, r is the remainder and b is the divisor.

Putting the values in equation (1) -

⇒ 1575 = 960(1) + 615.

Here 615 is the remainder, which is not zero. Again applying the Euclid’s divison lemma -

Now, a = 960; b = 615;

⇒ 960 = 615(1) + 345.

Here 345 is the remainder, which is not zero. Again applying the Euclid’s divison lemma -

Now, a = 615; b = 345;

⇒ 615 = 345(1) + 270.

Here 270 is the remainder, which is not zero. Again applying the Euclid’s divison lemma -

Now, a = 345; b = 270;

⇒ 345 = 270(1) + 75.

Here 75 is the remainder, which is not zero. Again applying the Euclid’s divison lemma -

Now, a = 270; b = 75;

⇒ 270 = 75(3) + 45.

Here 45 is the remainder, which is not zero. Again applying the Euclid’s divison lemma -

Now, a = 75; b = 45;

⇒ 75 = 45(1) + 30.

Here 30 is the remainder, which is not zero. Again applying the Euclid’s divison lemma -

Now, a = 45; b = 30;

⇒ 45 = 30(1) + 15.

Here 15 is the remainder, which is not zero. Again applying the Euclid’s divison lemma -

a = 30; b = 15;

⇒ 30 = 15(2) + 0.

∵ remainder is zero.

∴ HCF is 15.