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Question Number 182272 by SEKRET last updated on 06/Dec/22

1≤a≤37 1≤b≤37 1+7a+8b +19ab = 0 mod(37) a and b natural nambers (a_1 ;b_1 ) (a_2 ;b_2 )......(a_n ;b_n ) n=?

$$\:\:\:\:\mathrm{1}\leqslant\boldsymbol{\mathrm{a}}\leqslant\mathrm{37} \\ $$$$\:\:\:\:\mathrm{1}\leqslant\boldsymbol{\mathrm{b}}\leqslant\mathrm{37} \\ $$$$\:\:\:\mathrm{1}+\mathrm{7a}+\mathrm{8}\boldsymbol{\mathrm{b}}\:+\mathrm{19}\boldsymbol{\mathrm{ab}}\:\:=\:\mathrm{0}\:\boldsymbol{\mathrm{mod}}\left(\mathrm{37}\right)\: \\ $$$$\:\:\boldsymbol{\mathrm{a}}\:\:\:\boldsymbol{\mathrm{and}}\:\:\boldsymbol{\mathrm{b}}\:\:\boldsymbol{\mathrm{natural}}\:\:\boldsymbol{\mathrm{nambers}} \\ $$$$\:\:\:\:\:\left(\boldsymbol{\mathrm{a}}_{\mathrm{1}} ;\boldsymbol{\mathrm{b}}_{\mathrm{1}} \right)\:\left(\boldsymbol{\mathrm{a}}_{\mathrm{2}} ;\boldsymbol{\mathrm{b}}_{\mathrm{2}} \right)......\left(\boldsymbol{\mathrm{a}}_{\boldsymbol{\mathrm{n}}} ;\boldsymbol{\mathrm{b}}_{\boldsymbol{\mathrm{n}}} \right) \\ $$$$\:\:\:\boldsymbol{\mathrm{n}}=? \\ $$$$\:\:\:\: \\ $$

Commented by SEKRET last updated on 07/Dec/22

?

$$\: \\ $$$$\:\:\:\:? \\ $$

Answered by nikif99 last updated on 07/Dec/22

n=73

$${n}=\mathrm{73} \\ $$

Commented by SEKRET last updated on 07/Dec/22

?

$$? \\ $$

Commented by SEKRET last updated on 07/Dec/22

good ?

$$\:\boldsymbol{\mathrm{good}}\:\:\:? \\ $$

Answered by Rasheed.Sindhi last updated on 07/Dec/22

1≤a≤37 ; 1≤b≤37 ; a,b∈N 1+7a+8b +19ab = 0 mod(37) (a_1 ;b_1 ) (a_2 ;b_2 )......(a_n ;b_n ) ; n=?_(−^ ) 7a+8b +19ab+1+37ab ≡ 0 mod(37) 7a +56ab+8b+1 ≡0 mod(37) 7a(1+8b)+(1+8b) ≡ 0 mod(37) (7a+1)(8b+1)≡0 mod(37) (7a+1)(8b+1)≡37k mod(37) 7a+1=37 ∧ 8b+1=k mod(37) 7a≡36+3(37) ∧ 8b≡k−1 mod(37) a=21 ∧ b=1,2,...,37 ( 37 pairs) 8b+1≡37 ∧ 7a+1≡k mod(37) 8b≡36+4(37) ∧ 7a≡k−1mod(37) b≡23 ∧ a=1,2,...,37 (37 pairs) Common pair (21,23) Total pairs=37×2−1=73 n=73

$$\:\:\:\:\mathrm{1}\leqslant\boldsymbol{\mathrm{a}}\leqslant\mathrm{37}\:\:;\:\:\:\mathrm{1}\leqslant\boldsymbol{\mathrm{b}}\leqslant\mathrm{37}\:;\:\mathrm{a},\mathrm{b}\in\mathbb{N} \\ $$$$\:\:\:\mathrm{1}+\mathrm{7a}+\mathrm{8}\boldsymbol{\mathrm{b}}\:+\mathrm{19}\boldsymbol{\mathrm{ab}}\:\:=\:\mathrm{0}\:\boldsymbol{\mathrm{mod}}\left(\mathrm{37}\right)\: \\ $$$$\underset{−^{} } {\:\:\:\left(\boldsymbol{\mathrm{a}}_{\mathrm{1}} ;\boldsymbol{\mathrm{b}}_{\mathrm{1}} \right)\:\left(\boldsymbol{\mathrm{a}}_{\mathrm{2}} ;\boldsymbol{\mathrm{b}}_{\mathrm{2}} \right)......\left(\boldsymbol{\mathrm{a}}_{\boldsymbol{\mathrm{n}}} ;\boldsymbol{\mathrm{b}}_{\boldsymbol{\mathrm{n}}} \right)\:;\:\mathrm{n}=?} \\ $$$$\:\:\:\mathrm{7a}+\mathrm{8}\boldsymbol{\mathrm{b}}\:+\mathrm{19ab}+\mathrm{1}+\mathrm{37ab}\:\:\equiv\:\mathrm{0}\:\boldsymbol{\mathrm{mod}}\left(\mathrm{37}\right)\: \\ $$$$\:\:\:\mathrm{7a}\:+\mathrm{56ab}+\mathrm{8b}+\mathrm{1}\:\equiv\mathrm{0}\:\boldsymbol{\mathrm{mod}}\left(\mathrm{37}\right)\: \\ $$$$\:\:\:\mathrm{7a}\left(\mathrm{1}+\mathrm{8b}\right)+\left(\mathrm{1}+\mathrm{8b}\right)\:\equiv\:\mathrm{0}\:\boldsymbol{\mathrm{mod}}\left(\mathrm{37}\right) \\ $$$$\left(\mathrm{7a}+\mathrm{1}\right)\left(\mathrm{8b}+\mathrm{1}\right)\equiv\mathrm{0}\:\boldsymbol{\mathrm{mod}}\left(\mathrm{37}\right) \\ $$$$\left(\mathrm{7a}+\mathrm{1}\right)\left(\mathrm{8b}+\mathrm{1}\right)\equiv\mathrm{37k}\:\boldsymbol{\mathrm{mod}}\left(\mathrm{37}\right) \\ $$$$\mathrm{7a}+\mathrm{1}=\mathrm{37}\:\wedge\:\mathrm{8b}+\mathrm{1}=\mathrm{k}\:\boldsymbol{\mathrm{mod}}\left(\mathrm{37}\right) \\ $$$$\mathrm{7a}\equiv\mathrm{36}+\mathrm{3}\left(\mathrm{37}\right)\:\wedge\:\mathrm{8b}\equiv\mathrm{k}−\mathrm{1}\:\boldsymbol{\mathrm{mod}}\left(\mathrm{37}\right) \\ $$$$\mathrm{a}=\mathrm{21}\:\wedge\:\mathrm{b}=\mathrm{1},\mathrm{2},...,\mathrm{37}\:\left(\:\mathrm{37}\:{pairs}\right) \\ $$$$\mathrm{8b}+\mathrm{1}\equiv\mathrm{37}\:\wedge\:\mathrm{7a}+\mathrm{1}\equiv\mathrm{k}\:\:\boldsymbol{\mathrm{mod}}\left(\mathrm{37}\right) \\ $$$$\mathrm{8b}\equiv\mathrm{36}+\mathrm{4}\left(\mathrm{37}\right)\:\wedge\:\mathrm{7a}\equiv\mathrm{k}−\mathrm{1}\boldsymbol{\mathrm{mod}}\left(\mathrm{37}\right) \\ $$$$\mathrm{b}\equiv\mathrm{23}\:\wedge\:\mathrm{a}=\mathrm{1},\mathrm{2},...,\mathrm{37}\:\left(\mathrm{37}\:{pairs}\right) \\ $$$$\mathrm{Common}\:\mathrm{pair}\:\left(\mathrm{21},\mathrm{23}\right) \\ $$$$\mathcal{T}{otal}\:{pairs}=\mathrm{37}×\mathrm{2}−\mathrm{1}=\mathrm{73} \\ $$$${n}=\mathrm{73} \\ $$

Commented by SEKRET last updated on 07/Dec/22

thank you

$$\:\boldsymbol{\mathrm{thank}}\:\:\mathrm{you}\: \\ $$

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