How-many-ordered-pairs-m-n-are-there-for-1-m-n-100-such-that-7-m-7-n-is-divisble-by-5-

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} \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, \dots)} \equiv (1, 2, 4, 3, 1, 2, 4, \dots) (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 .