Solve-the-following-7-x-13-mod-18-Pl-give-complete-process-

Question Number 13744 by AH Soomro last updated on 23/May/17

S o l v e t h e f o l l o w i n g

7^{x} \equiv 13 (m o d 18)

P l g i v e c o m p l e t e p r o c e s s .

Commented by prakash jain last updated on 23/May/17

7 \equiv 7 (\mod 18)

7^{2} = 49 \equiv 13 (\mod 18)

7^{3} = 343 \equiv 1 (\mod 18)

7^{3 n} \equiv 1 (\mod 18)

7^{3 n + 1} \equiv 7 (\mod 18)

7^{3 n + 2} \equiv 13 (\mod 18)

x = 3 n + 2, n \in {0, 1, 2 \dots}

Commented by mrW1 last updated on 23/May/17

\oint a n t a s t i C!

Commented by RasheedSindhi last updated on 24/May/17

We can come directly to

third step, because {Euler}^{'} s

theorem tells us that

If (a, m) = 1, then

a^{\emptyset (m)} \equiv 1 (\mod m)

[where \emptyset is Euler function

and \emptyset (m) is meant number of

coprimes of m which are not

greater than m .]

Hence 7^{3} \equiv 1 (\mod 18)

∵ \emptyset (7) = 3 and (7, 18) = 1

Commented by AH Soomro last updated on 24/May/17

Thanks to all .

Answered by mrW1 last updated on 23/May/17

7^{x} = 18 n + 13 = 7 \times 2 (n + 1) + 4 n - 1

4 n - 1 = 7 m

n = \frac{7 m + 1}{4}

\Rightarrow 7^{x} = 18 n + 13 = \frac{18 \times (7 m + 1) + 52}{4} = \frac{7 (9 m + 5)}{2}

\frac{9 m + 5}{2} = 7^{i}

m = \frac{2 \times 7^{i} - 5}{9}

f o r m t o b e i n t e g e r \Rightarrow i = 1, 4, 7 \dots o r 3 k - 2 w i t h k \in N^{+}

i . e . m = \frac{2 \times 7^{3 k - 2} - 5}{9} (* s e e c o m m e n t)

\Rightarrow 7^{x} = \frac{7 (18 m + 5)}{2} = 7 \times 7^{3 k - 2} = 7^{3 k - 1}

\Rightarrow S o l u t i o n i s x = 3 k - 1, k \in N^{+}

i . e . x = 2, 5, 8, 11 \dots

Commented by mrW1 last updated on 23/May/17

H e r e t h e p r o o f t h a t

m = \frac{2 \times 7^{3 k - 2} - 5}{9} = i n t e g e r

o r

2 \times 7^{3 k - 2} - 5 i s d i v i s i b l e b y 9 .

(1) c h e c k f o r k = 1

2 \times 7^{3 \times 1 - 2} - 5 = 2 \times 7 - 5 = 9 \equiv 0 (m o d 9)

\Rightarrow t r u e

(2) s u p p o s e d {i t}^{'} s t r u e f o r k

i . e . 2 \times 7^{3 k - 2} - 5 \equiv 0 (m o d 9)

f o r k + 1 w e h a v e :

2 \times 7^{3 (k + 1) - 2} - 5

= 2 \times 7^{3} \times 7^{3 k - 2} - 7^{3} \times 5 + 7^{3} \times 5 - 5

= 7^{3} \times (2 \times 7^{3 k - 2} - 5) + (7^{3} - 1) \times 5

= 7^{3} \times (2 \times 7^{3 k - 2} - 5) + 38 \times 9 \times 5 \equiv 0 (m o d 9)

\Rightarrow {i t}^{'} s t r u e f o r k + 1

p r o v e d!

Commented by AH Soomro last updated on 24/May/17

T h a n k s .