Find whether the pair of numbers 615, 154 is co-prime or not.
Answers
Step-by-step explanation:
Given, the two numbers 847,2160
Here, Co-primes are 2 numbers which have only one as a common factor.
Let, a=2160
b=847
Then, by Euclid's lemma
a=bq+r,0≤r<b
So,
2160=847×2+466
847=466×1+381
466=381×1+85
381=85×4+41
85=41×2+3
41=3×13+2
3=2×1+1
2=1×2+0
As 1 is the H.C.F of 847 and 2160,
∴ 847 and 2160 are co-primes as they have only 1 as their H.C.F.
Answer:
Givennumbers615and154are
\green { Co-prime }Co−prime
Step-by-step explanation:
\underline { \pink { Co-prime }}
Co−prime
\begin{gathered} If \: HCF \: of \: two \: numbers \: equal \: to \: 1 \\ then \: they \: are \: called \: Co-prime. \end{gathered}
IfHCFoftwonumbersequalto1
thentheyarecalledCo−prime.
Given \: numbers \: 615 \: and \: 154Givennumbers615and154
Here , 615 > 154Here,615>154
\begin{gathered} Start \: with \:the \: larger \:integer \: , that \:is, \\615. Apply \: the \: Euclid's \: division \: algorithm \\we \:get \end{gathered}
Startwiththelargerinteger,thatis,
615.ApplytheEuclid
′
sdivisionalgorithm
weget
615 = 154 \times 3 + 53615=154×3+53
\begin{gathered} Since, \: the \: remainder \: 53 ≠ 0 , we \\apply \: the \: division \: algorithm \:to \: 53, to \:get \end{gathered}
Since,theremainder53
=0,we
applythedivisionalgorithmto53,toget
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}
Theremainderhasnowbecome
zero,soourprocedurestops.
\begin{gathered} Since, the \: divisor \: at \:this \:stage \:is \: 1,\\HCF \: of \: 615\: and \: 154 \: is \: 1 . \end{gathered}
Since,thedivisoratthisstageis1,