Given a, b ∈[0;4] 309a+15c=226b 1) show that 2b≡0[3] and 3a≡1[5]


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Question Number 122358 by mathocean1 last updated on 16/Nov/20

$G i v e n a, b \in [0; 4]$ $309 a + 15 c = 226 b$ $1) s h o w t h a t 2 b \equiv 0 [3] a n d 3 a \equiv 1 [5]$
	Answered by mindispower last updated on 16/Nov/20

	$309 a + 15 c = 226 b = 225 b + b$ $309 a + 15 c = 3 (103 a + 5 c) \equiv 0 [3]$ $\Rightarrow 3 (75 b) + b \equiv 0 [3] \Leftrightarrow b \equiv 0 [3] \Rightarrow 2 b = 0.2 [3] \equiv 0 [3]$ $309 a = 226 b - 15 c \equiv b [5]$ $309 = 4 + 5.61$ $\Leftrightarrow 4 a \equiv b [5] \Rightarrow - a = b [5] \Rightarrow a = - b [5]$ $\Rightarrow 3 a = 2 b [5]$ $2 b = 3 k f i r s t r e s u l t, \Rightarrow 2 b \in {0, 6}$ $b = 0 \Rightarrow 309.0 + 15.0 = 0, {0, 0, 0} s o l u t i o n$ $3 a \neq 1 [5]$ $b \neq 0 \Rightarrow 2 b = 6 \equiv 1 [5]$ $3 a = 2 b [5] \Leftrightarrow 3 a = 1 [5], w e m u s t e b \in [1, 4]$
	Commented by mathocean1 last updated on 18/Nov/20

	$t h a n k s s i r$
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