(i) To find the H.C.F. of 340 and 412. Using Euclid’s division algorithm.
We get 412 = 340 x 1 + 72
The remainder 72 ≠ 0
Again applying Euclid’s division algorithm
340 = 72 x 4 + 52
The remainder 52 ≠ 0.
Again applying Euclid’s division algorithm
72 = 52 x 1 + 20
The remainder 20 ≠ 0.
Again applying Euclid’s division algorithm,
52 = 20 x 2 + 12
The remainder 12 ≠ 0.
Again applying Euclid’s division algorithm.
20 = 12 x 1 + 8
The remainder 8 ≠ 0.
Again applying Euclid’s division algorithm
12 = 8 x 1 + 4
The remainder 4 ≠ 0.
Again applying Euclid’s division algorithm
8 = 4 x 2 + 0
The remainder is zero.
Therefore H.C.F. of 340 and 412 is 4.
(ii) To find the H.C.F. of 867 and 255, using Euclid’s division algorithm.
867 = 255 x 3 + 102
The remainder 102 ≠ 0.
Again using Euclid’s division algorithm
255 = 102 x 2 + 51
The remainder 51 ≠ 0.
Again using Euclid’s division algorithm
102 = 51 x 2 + 0
The remainder is zero.
Therefore the H.C.F. of 867 and 255 is 51.
(iii) To find H.C.F. 10224 and 9648. Using Euclid’s division algorithm.
10224 = 9648 x 1 + 576
The remainder 576 ≠ 0.
Again using Euclid’s division algorithm
9648 = 576 x 16 + 432
Remainder 432 ≠ 0.
Again applying Euclid’s division algorithm
576 = 432 x 1 + 144
Remainder 144 ≠ 0.
Again using Euclid’s division algorithm
432 = 144 x 3 + 0
The remainder is zero.
There H.C.F. of 10224 and 9648 is 144.
(iv) To find H.C.F. of 84, 90 and 120.
Using Euclid’s division algorithm
90 = 84 × 1 + 6
The remainder 6 ≠ 0.
Again using Euclid’s division algorithm
84 = 6 × 14 + 0
The remainder is zero.
∴ The H.C.F. of 84 and 90 is 6. To find the H.C.F. of 6 and 120 using Euclid’s division algorithm.
120 = 6 × 20 + 0
The remainder is zero.
Therefore H.C.F. of 120 and 6 is 6
∴ H.C.F. of 84, 90 and 120 is 6
