Question

Using Euclid division algorithm find which of the following pairs of number are coprime question 615 154

Answer 1

Step-by-step explanation:

Givennumbers615and154are

Step-by-step explanation:

If \: HCF \: of \: two \: numbers \: equal \: to \: 1 \\ then \: they \: are \: called \: Co-prime. \end{gathered}

If HCF of two numbers equal to 1

then they are called Co−prime.

Given \: numbers \: 615 \: and \: 154Givennumbers615and154

Here , 615 > 154Here,615>154

Start \: with \:the \: larger \:integer \: , that \:is, \\615. Apply \: the \: Euclid's \: division \: algorithm \\we \:get \end{gathered}

Start with the larger integer,that is,

615.Apply the Euclid

′

s division algorithm

we get

615 = 154 \times 3 + 53615=154×3+53

Since, \: the \: remainder \: 53 ≠ 0 , we \\apply \: the \: division \: algorithm \:to \: 53, to \:get \end{gathered}

Since,the remainder is 53



=0,we

apply the division algorithm to 53,to get

154 = 53 \times 2 + 48154=53×2+48

53 = 48 \times 1 + 553=48×1+5

48 = 5 \times 9 + 348=5×9+3

5 = 3\times 1 + 25=3×1+2

3 = 2\times 1 + 13=2×1+1

2 = 1\times 2 + 02=1×2+0

\begin{gathered} The \: remainder \: has \: now \: become \\zero, \: so \: our \: procedure \: stops . \end{gathered}

The remainder has now become

zero,so our procedure stops.

Since, the \: divisor \: at \:this \:stage \:is \: 1,\\HCF \: of \: 615\: and \: 154 \: is \: 1 . \end{gathered}

Since,the divisor at this stage is 1,

HCFof615and154is1.

Therefore.,

Given \: numbers \:615 \: and \: 154 \: are }Given numbers 615and154are

Co−prime