Given a = 1+3+3^2 +3^3 +3^4 +...+3^(100) Find the remainder of dividing the number by 5 . (a) 2 (b) 0 (c)4 (d)1 (e) 3


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Question Number 119956 by bobhans last updated on 28/Oct/20

$G i v e n a = 1 + 3 + 3^{2} + 3^{3} + 3^{4} + . . . + 3^{100}$ $F i n d t h e r e m a i n d e r o f d i v i d i n g t h e n u m b e r$ $b y 5 .$ $(a) 2 (b) 0 (c) 4 (d) 1 (e) 3$
	Answered by talminator2856791 last updated on 08/Sep/21

	$a = \underset{k = 0}{\sum^{24}} 3^{4 k} (1 + 3^{2}) + \underset{k = 0}{\sum^{24}} 3^{4 k + 1} (1 + 3^{2}) + 3^{100}$ $\underset{k = 0}{\sum^{24}} 3^{4 k} (1 + 3^{2}) = 10 \underset{k = 0}{\sum^{24}} 3^{4 k}$ $\underset{k = 0}{\sum^{24}} 3^{4 k} (1 + 3^{2}) = 10 \underset{k = 0}{\sum^{24}} 3^{4 k + 1}$ $\to 5 ∣ 10 \underset{k = 0}{\sum^{24}} 3^{4 k}, 5 ∣ 10 \underset{k = 0}{\sum^{24}} 3^{4 k + 1}$ $\to 5 ∣ 1 + 3 + 3^{2} + . . . . . + 3^{99}$ $\Rightarrow 5 ∣ 3 (1 + 3 + 3^{2} + . . . . . + 3^{99})$ $= 5 ∣ 3 + 3^{2} + 3^{3} + . . . . . + 3^{100}$ $\Rightarrow 1 + 3 + 3^{2} + . . . . . . + 3^{100} \equiv 1 (\mod 5)$ $answer : d$
	Answered by TANMAY PANACEA last updated on 28/Oct/20

	$S = \frac{3^{101} - 1}{3 - 1} = \frac{1}{2} (3^{101} - 1)$ $3^{1} = 3$ $3^{2} = 9$ $3^{3} = 27$ $3^{4} = 81$ $3^{5} = 243$ $3^{6} = 729$ $s o l a s t d i g i t r e p e t s (3 \to 9 \to 7 \to 1)$ $\frac{101}{4} \to 25 c y c l e o f (3 \to 9 \to 7 \to 1) + p l u s 1$ $l a s t d i g i t o f 3^{101} i s 3$ $(n o w l a s t d i g i t 3) - 1$ $s o l a s t d i g i t b e c o m e s = 2$ $w h e n d e v i d e d b y 2$ $l a s t d i g i t b e c o m e s 1$ $\frac{3^{101} - 1}{2} = l a s t d i g i t = 1$ $\frac{3^{101} - 1}{2} = (5 \times e v e n m u l t i p l e) + l a s t d i g i t = 1$
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