Question

I need help with this question on Discrete Mathematics with Applications

a. 8. Solve the following equivalences, if it cannot be solved, explain why not: 3x – 7 = 5(mod 7) b. 4x + 12 = 18(mod 24) C. 5x – 11 = 9(mod 13)

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Answer 1

ffa and n are coprimes, than lineas congmence axa bimod n ) has solutions that are given by.. x= vb (mod n) where v is any multiplicative envere da medulon asx 3x - 7=5(mod 7). first convert above into ax=b (modn) form 3x+7-547K 3X-7=5+7k 3X= 5+7K+7 3*+ 372 7k+ 3X= 12+7K 3X2 12 (mad 7 ) 3X512 (moda) => 3x = 5 (mod)) 3 and all coprenes. multiplicature inverse of 3 and 7 is s 5 SOX= 5(12) (mod 7) Go (modal ANS x= Go (moda? bl: 4X+12= 18(mod 24) Convert into ax= bemodni formi 4X+12= 18+24k 4X= 18+246-12 463 6+ 24K => 4X = mod 24) - 4X=6(mod24]

The lineas congruent has a solution If highest common factor of 4 and 24 i.e: 49 and 6 is not divusible. by 4. ANS so 4X+12E 18(mod 247 has no solutions c): = 5X-11 = q (mod 13): Convert into ax=b (modn) form. 5X-11 = 9+13k SX = 9+13k +1 Si= 13k+20) SX-20 +13k => 5X = 20 (mod 13) Here sand 13 are coprimes. mallinticative inverse of and 13 is 8 80 x= (g/b final n) X= 8(2016 med x= 160 moda ANS x= 160 (mod 13)

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