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Question Number 57602 by MJS last updated on 08/Apr/19

Commented by MJS last updated on 08/Apr/19

$reposted qu . 21795$
	Answered by MJS last updated on 08/Apr/19
	$it′s easy the remainders of (2^x /7) with x∈N: {1, 2, 4, 1, 2, 4, ...} the remainders of (3^x /7) with x∈N: {1, 3, 2, 6, 4, 5, 1, 3, 2, 6, 4, 5, ...} the sum of these: {2, 5, 6, 7, 6, 2, 2, 5, 6, 7, 6, 2, ...} ⇒ x=3+6n$
	${it}^{'} s easy$ $the remainders of \frac{2^{x}}{7} with x \in N :$ ${1, 2, 4, 1, 2, 4, . . .}$ $the remainders of \frac{3^{x}}{7} with x \in N :$ ${1, 3, 2, 6, 4, 5, 1, 3, 2, 6, 4, 5, . . .}$ $the sum of these :$ ${2, 5, 6, 7, 6, 2, 2, 5, 6, 7, 6, 2, . . .}$ $\Rightarrow x = 3 + 6 n$
	Commented by mr W last updated on 08/Apr/19

	$v e r y n i c e a p p r o a c h s i r!$
	Commented by MJS last updated on 08/Apr/19

	$thank you . I was just trying . . .$
	Commented by mr W last updated on 09/Apr/19

	$i g o t t h e s a m e r e s u l t x = 3 (2 k + 1), b u t$ $i w a s n o t s u r e i f t h e r e a r e o t h e r s o l u t i o n s .$
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