How many ordered pairs (m,n) are there for 1≤m,n≤100 such that 7^m +7^n is divisble by 5?


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Question Number 2878 by Yozzi last updated on 29/Nov/15

$H o w m a n y o r d e r e d p a i r s (m, n) a r e t h e r e$ $f o r 1 ⩽ m, n ⩽ 100 s u c h t h a t 7^{m} + 7^{n}$ $i s d i v i s b l e b y 5 ?$
Commented by prakash jain last updated on 29/Nov/15

$I read the statment 1 ⩽ m, n ⩽ 100 as 2 separate$ $statement . Will update the answer .$ $1 ⩽ m$ $n ⩽ 100$
	Answered by prakash jain last updated on 29/Nov/15
	$7^m +7^n ≡0 (mod 5) 2^m +2^n ≡0 (mod 5) k∈{0,1,2,3} k=4j⇒2^k ≡1 (mod 5) k=4j+1⇒2^k ≡2 (mod 5) k=4j+2⇒2^k ≡4 (mod 5) k=4j+3⇒2^k ≡3 (mod 5) Solution for m and n j,l∈{0,1,2,3} m=4j, n=4l+2 m=4j+1, n=4l+3 m=4j+2, n=4l m=4j+3, n=4l+1 1≤n≤100, 1≤m≤100 n=4l 1≤l≤25 m=4j+2 0≤j≤24 n=4l+1 0≤l≤24 m=4j+3, 0≤j≤24 n=4l+2 0≤l≤24 m=4j 1≤j≤25 n=4l+3 0≤l≤24 m=4j+1 0≤j≤24 Total number of pairs=4×25^2 =625×4=2500$
	$7^{m} + 7^{n} \equiv 0 (\mod 5)$ $2^{m} + 2^{n} \equiv 0 (\mod 5)$ $k \in {0, 1, 2, 3}$ $k = 4 j \Rightarrow 2^{k} \equiv 1 (\mod 5)$ $k = 4 j + 1 \Rightarrow 2^{k} \equiv 2 (\mod 5)$ $k = 4 j + 2 \Rightarrow 2^{k} \equiv 4 (\mod 5)$ $k = 4 j + 3 \Rightarrow 2^{k} \equiv 3 (\mod 5)$ $Solution for m and n$ $j, l \in {0, 1, 2, 3}$ $m = 4 j, n = 4 l + 2$ $m = 4 j + 1, n = 4 l + 3$ $m = 4 j + 2, n = 4 l$ $m = 4 j + 3, n = 4 l + 1$ $1 ⩽ n ⩽ 100, 1 ⩽ m ⩽ 100$ $n = 4 l 1 ⩽ l ⩽ 25$ $m = 4 j + 2 0 ⩽ j ⩽ 24$ $n = 4 l + 1 0 ⩽ l ⩽ 24$ $m = 4 j + 3, 0 ⩽ j ⩽ 24$ $n = 4 l + 2 0 ⩽ l ⩽ 24$ $m = 4 j 1 ⩽ j ⩽ 25$ $n = 4 l + 3 0 ⩽ l ⩽ 24$ $m = 4 j + 1 0 ⩽ j ⩽ 24$ $Total number of pairs = 4 \times 25^{2} = 625 \times 4 = 2500$
	Answered by Rasheed Soomro last updated on 30/Nov/15

	$L e t 7^{m} \equiv r (m o d 5)$ $E x p e r i m e n t i n g a n d O b s e r v i n g f o r v a l u e s o f m a n d r :$ $7^{(0, 1, 2, 3, 4, 5, 6, . . .)} \equiv (1, 2, 4, 3, 1, 2, 4, . . .) (m o d 5)$ $G e n e r a l i z i n g :$ $7^{4 k} \equiv 1 (m o d 5) ∥ 7^{4 h + 2} \equiv 4 (m o d 5)$ $7^{4 k + 1} \equiv 2 (m o d 5) ∥ 7^{4 h + 3} \equiv 3 (m o d 5)$ $7^{4 k + 2} \equiv 4 (m o d 5) ∥ 7^{4 h} \equiv 1 (m o d 5)$ $7^{4 k + 3} \equiv 3 (m o d 5) ∥ 7^{4 h + 1} \equiv 2 (m o d 5)$ $A d d i n g c o r r e s p o n d i n g$ $7^{4 k} + 7^{4 h + 2} \equiv 0 (m o d 5)$ $7^{4 k + 1} + 7^{4 h + 3} \equiv 0 (m o d 5)$ $7^{4 k + 2} + 7^{4 h} \equiv 0 (m o d 5)$ $7^{4 k + 3} + 7^{4 h + 1} \equiv 0 (m o d 5)$ $F o u r t y p e s o f o r d e r e d p a i r s [(m, n)] s a t i s f y i n g t h e$ $g i v e n s t a t e m e n t .$ $1 ⩽ (4 k, 4 h + 2) ⩽ 100$ $1 ⩽ (4 k + 1, 4 h + 3) ⩽ 100$ $1 ⩽ (4 k + 2, 4 h) ⩽ 100$ $1 ⩽ (4 k + 3, 4 h + 1) ⩽ 100$ $N o w {i t}^{'} s e a s y t o c o u n t$ $S u p p o s e t h e r e a r e x 4 k + 1 t y p e n u m b e r s b e t w e e n 1 a n d$ $100 i n c l u s i v e a n d y 4 h + 3 t y p e n u m b e r s b e t w e e n 1 a n d$ $100 i n c l u s i v e THEN$ $T h e r e a r e x y o r d e r e d p a i r s o f s u c h n u m b e r s .$ $A m u l t i p l i c a t i o n a l t a b l e i s e a s y w a y o f r e c o r d i n g a n d$ $c o u n t i n g a l l r e q u i r e d {(m, n)}^{'} s$
	Commented by prakash jain last updated on 29/Nov/15

	$Thanks .$ $I thought$ $1 ⩽ m$ $n ⩽ 100$
	Commented by Rasheed Soomro last updated on 29/Nov/15

	$I t^{'} s n o t m i s r e a d i n g . A c t u a l l y t h e s t a t e m e n t c a n a l s o b e$ $r e a d i n t h a t w a y .$
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