Question

Could anyone help me?

Answer 1

this is your answer

do next by your own

in. B part

firstly u have to find hcf of first two

then that of last two

Answer 2

Answer:

$1. \\ consider \: 256 \: 352 \\ a = 352 \\ b = 256 \\ by \: euclids \: division \: of \: algorithm \\ a = bq + r \\ 0 < or = r < b \\ \\ 352 = 256 \times 1 + 96 \\ 256 = 96 \times 2 + 64 \\ 96 = 64 \times 1 + 32 \\ 64 = 32 \times 2 + 0 \\ \\ therefore \: we \: get \: \\ HCF \: of \: {256 \: and \: 352} \: is \: 32$

hope u hav understood ✌☺

$2. \\cosider \: 450 \: 500 \: 625 \\ first \: we \: take \: \\ a = 625 \\ b = 500 \\ soo \: frm \: eculids \: division \: \\ a = bq + r \\ 0 < or = < b \\ 625 = 500 \times 1 + 125 \\ 500 = 125 \times 4 + 0 \\ terefore \: HCF \: of \: 625 \: and \: 500 = 125 \\ \\ next \: \: we \: take \\ a = 450 \\ b = 125 \\ 450 = 125 \times 3 + 75 \\ 125 = 75 \times 1 + 50 \\ 75 = 50 \times 1 + 25 \\ 50 = 25 \times 2 + 0 \\ therefore \: HCF \: of \: 450, \: 500, \: and \: 625 = 25$

hope u hav understood ✌☺