Solve 21^a + 28^b = 35^c if a, b, and c are positive integers.


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Question Number 42908 by naka3546 last updated on 04/Sep/18

$S o l v e$ $21^{a} + 28^{b} = 35^{c}$ $i f$ $a, b, a n d c a r e p o s i t i v e i n t e g e r s .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 04/Sep/18

	$l e t m e t r y b y l o g i c$ $35^{c} = t h e l a s t d i g i t i s 5$ $21^{a} = t h e l a s t d i g i t i s 1$ $s o l a s t d i g i t o f 28^{b} s h o u l d b e 4 s o b i s m u l t i p l e$ $2 t h a t i s b = 2 k$ $21^{a} + 28^{2 k} = 35^{c}$ $l e t t r i a l c = 2 35^{2} = 1225$ $28^{2 \times 1} = 784$ $n o w 21^{a} = 1225 - 784 = 441 = 21^{2}$ $h e n c e a = 2 b = 2 c = 2$
	Commented by MJS last updated on 04/Sep/18

	$solution is right but the sequence of the$ $last digit of 28^{b} is$ $8 4 2 6 8 4 . . . \Rightarrow b = 2 + 4 k$
	Commented by tanmay.chaudhury50@gmail.com last updated on 05/Sep/18

	$28^{1} = l a s t d i g i t 8 y e s s i r$ $28^{2} = l a s t d i g i t 4$ $28^{3} = l a s t d i g i t 2$ $28^{4} = l a s t d i g i t 6$ $28^{5} = l a s t d i g i t 8$ $s o s e q u e n c e i s 8 4 2 6 8 4 2 6$
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