Given-that-7-k-1-mod-15-a-Write-down-three-values-of-k-b-Find-the-general-solution-of-the-equation-7-k-1-mod-15-

Question Number 80505 by Rio Michael last updated on 03/Feb/20

Given that 7^{k} \equiv 1 (\mod 15)

a) Write down three values of k .

b) Find the general solution of

the equation 7^{k} \equiv 1 (m o d 15)

Commented by mr W last updated on 03/Feb/20

k = 4 n w i t h n \in N

Commented by Rio Michael last updated on 03/Feb/20

how sir ?

Commented by mr W last updated on 03/Feb/20

7^{k} = 15 m + 1

l a s t d i g i t o f 7^{k} c a n b e 7, 9, 3, 1

l a s t d i g i t o f 15 m + 1 c a n b e 6, 1

t h e o n l y p o s s i b i l i t y i s t h a t b o t h h a v e

t h e l a s t d i g i t 1 . t h a t m e a n s 7^{k} = {(7^{2} \times 7^{2})}^{n},

i . e . k = 4 n . w e c a n p r o v e t h a t 7^{4 n} - 1

i s i n d e e d a m u l t i p l e o f 15 .

Commented by Rio Michael last updated on 03/Feb/20

s i r h o w d o y o u g e t t h i e r l a s t

d i g i t s p l e a s e

Commented by mr W last updated on 03/Feb/20

7 \Rightarrow 7

7 \times 7 \Rightarrow 9

7 \times 7 \times 7 \Rightarrow 3

7 \times 7 \times 7 \times 7 \Rightarrow 1

7 \times 7 \times 7 \times 7 \times 7 \Rightarrow 7 e t c .

15 \times 1 + 1 \Rightarrow 6

15 \times 2 + 1 \Rightarrow 1

15 \times 3 + 1 \Rightarrow 6 e t c .

Commented by mr W last updated on 04/Feb/20

h e r e t h e p r o o f t h a t 7^{4 n} \equiv 1 m o d (15)

7^{4 n} = 2401^{n} = {(2400 + 1)}^{n} = \underset{r = 0}{\sum^{n}} C_{r}^{n} \times 2400^{r}

= 1 + \underset{r = 1}{\sum^{n}} C_{r}^{n} \times 2400^{r}

= 1 + \underset{r = 1}{\sum^{n}} C_{r}^{n} \times {(160 \times 15)}^{r}

\equiv 1 m o d (15)

Commented by Rio Michael last updated on 04/Feb/20

t h a n k s s o m u c h s i r

Commented by mr W last updated on 04/Feb/20

s e e a l s o Q 80580

Facebook Tweet Pin