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69x-1-mod-31-solve-for-x-




Question Number 96928 by bobhans last updated on 05/Jun/20
69x ≡ 1 (mod 31)   solve for x
69x1(mod31)solveforx
Commented by RAMANA last updated on 05/Jun/20
send the answer please
sendtheanswerplease
Answered by MAB last updated on 05/Jun/20
69x≡1[31]  69x−31×2×x≡1[31]  7x≡1[31]  we have 31 is prime and 7x≡1[31]   and the only inverse of 7 in the class 31  is 9 ( 7×9=63=2×31+1)  therfore x≡ [31]
69x1[31]69x31×2×x1[31]7x1[31]wehave31isprimeand7x1[31]andtheonlyinverseof7intheclass31is9(7×9=63=2×31+1)therforex[31]
Answered by 1549442205 last updated on 06/Jun/20
we have 69x−1=31a(a∈Z)⇔a=((69x−1)/(31))=2x+((7x−1)/(31))  ⇒7x−1=31b(b∈Z)⇔x=((31b+1)/7)=4b+((3b+1)/7)  ⇒3b+1=7c(c∈Z)⇔b=((7c−1)/3)=2c+((c−1)/3)  ⇒c−1=3d(d∈Z)⇒c=3d+1.From this we get  b=2c+d=7d+2⇒x=4b+c=31d+9  Thus,x=31d+9 or x=9(mod31)
wehave69x1=31a(aZ)a=69x131=2x+7x1317x1=31b(bZ)x=31b+17=4b+3b+173b+1=7c(cZ)b=7c13=2c+c13c1=3d(dZ)c=3d+1.Fromthiswegetb=2c+d=7d+2x=4b+c=31d+9Thus,x=31d+9orx=9(mod31)
Answered by mr W last updated on 06/Jun/20
69x=31n+1  69x−31n=1  ⇒x=31k+9 ⇒x≡9 mod 31    see also Q44704, Q19198
69x=31n+169x31n=1x=31k+9x9mod31seealsoQ44704,Q19198

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