x∈Z^+ 15^(48a+1) ≡ x (mod 17) find x=?


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Question Number 144961 by mathdanisur last updated on 30/Jun/21

$x \in Z^{+}$ $15^{48 a + 1} \equiv x (m o d 17) f i n d x = ?$
	Answered by Rasheed.Sindhi last updated on 01/Jul/21
	$15^8 ≡x(mod17) 15^8 ≡1(mod17) (15^8 )^(6a) ≡(1)^(6a) (mod17) 15^(48a) ≡1 15^(48a) .15≡1.15(mod17) 15^(48a+1) ≡15(mod17) x≡15(mod17) x=15+17k; k∈{0,1,2,....}$
	$15^{8} \equiv x (m o d 17)$ $15^{8} \equiv 1 (m o d 17)$ ${(15^{8})}^{6 a} \equiv {(1)}^{6 a} (m o d 17)$ $15^{48 a} \equiv 1$ $15^{48 a} . 15 \equiv 1.15 (m o d 17)$ $15^{48 a + 1} \equiv 15 (m o d 17)$ $x \equiv 15 (m o d 17)$ $x = 15 + 17 k; k \in {0, 1, 2, . . . .}$
	Commented by mathdanisur last updated on 01/Jul/21

	$c o o l S i r t h a n k y o u, a n s w e r x = 15 ?$
	Commented by Rasheed.Sindhi last updated on 01/Jul/21

	$15 + 17 k; W h e r e k \in {0, 1, 2, . . .}$
	Commented by mathdanisur last updated on 01/Jul/21

	$S i r, b u t a n s w e r : 15$
	Commented by Rasheed.Sindhi last updated on 01/Jul/21

	$T h e l e a s t p o s i t i v e i n t e g r a l v a l u e$ $o f x i s 15$
	Commented by mathdanisur last updated on 01/Jul/21

	$t h a n k s S i r$
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