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


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

$1 ⩽ a ⩽ 37$ $1 ⩽ b ⩽ 37$ $1 + 7 a + 8 b + 19 ab = 0 \mod (37)$ $a and b natural nambers$ $(a_{1}; b_{1}) (a_{2}; b_{2}) . . . . . . (a_{n}; b_{n})$ $n = ?$
Commented by SEKRET last updated on 07/Dec/22

$?$
	Answered by nikif99 last updated on 07/Dec/22

	$n = 73$
	Commented by SEKRET last updated on 07/Dec/22

	$?$
	Commented by SEKRET last updated on 07/Dec/22

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

	$1 ⩽ a ⩽ 37; 1 ⩽ b ⩽ 37; a, b \in N$ $1 + 7 a + 8 b + 19 ab = 0 \mod (37)$ $\underset{-^{}}{(a_{1}; b_{1}) (a_{2}; b_{2}) . . . . . . (a_{n}; b_{n}); n = ?}$ $7 a + 8 b + 19 ab + 1 + 37 ab \equiv 0 \mod (37)$ $7 a + 56 ab + 8 b + 1 \equiv 0 \mod (37)$ $7 a (1 + 8 b) + (1 + 8 b) \equiv 0 \mod (37)$ $(7 a + 1) (8 b + 1) \equiv 0 \mod (37)$ $(7 a + 1) (8 b + 1) \equiv 37 k \mod (37)$ $7 a + 1 = 37 \land 8 b + 1 = k \mod (37)$ $7 a \equiv 36 + 3 (37) \land 8 b \equiv k - 1 \mod (37)$ $a = 21 \land b = 1, 2, . . ., 37 (37 p a i r s)$ $8 b + 1 \equiv 37 \land 7 a + 1 \equiv k \mod (37)$ $8 b \equiv 36 + 4 (37) \land 7 a \equiv k - 1 \mod (37)$ $b \equiv 23 \land a = 1, 2, . . ., 37 (37 p a i r s)$ $Common pair (21, 23)$ $T o t a l p a i r s = 37 \times 2 - 1 = 73$ $n = 73$
	Commented by SEKRET last updated on 07/Dec/22

	$thank you$
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