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

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)

$$\mathrm{Given}\:\mathrm{that}\:\:\mathrm{7}^{{k}} \:\equiv\mathrm{1}\:\left(\mathrm{mod}\:\mathrm{15}\right) \\ $$$$\left.\mathrm{a}\right)\:\mathrm{Write}\:\mathrm{down}\:\mathrm{three}\:\mathrm{values}\:\mathrm{of}\:{k}. \\ $$$$\left.\mathrm{b}\right)\:\mathrm{Find}\:\mathrm{the}\:\mathrm{general}\:\mathrm{solution}\:\mathrm{of}\: \\ $$$$\mathrm{the}\:\mathrm{equation}\:\:\mathrm{7}^{{k}} \:\equiv\:\mathrm{1}\:\left({mod}\:\mathrm{15}\right) \\ $$

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

k=4n with n∈N

$${k}=\mathrm{4}{n}\:{with}\:{n}\in\mathbb{N} \\ $$

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

how sir?

$$\mathrm{how}\:\mathrm{sir}? \\ $$

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

7^k =15m+1 last digit of 7^k can be 7,9,3,1 last digit of 15m+1 can be 6,1 the only possibility is that both have the last digit 1. that means 7^k =(7^2 ×7^2 )^n , i.e. k=4n. we can prove that 7^(4n) −1 is indeed a multiple of 15.

$$\mathrm{7}^{{k}} =\mathrm{15}{m}+\mathrm{1} \\ $$$${last}\:{digit}\:{of}\:\mathrm{7}^{{k}} \:{can}\:{be}\:\mathrm{7},\mathrm{9},\mathrm{3},\mathrm{1} \\ $$$${last}\:{digit}\:{of}\:\mathrm{15}{m}+\mathrm{1}\:{can}\:{be}\:\mathrm{6},\mathrm{1} \\ $$$${the}\:{only}\:{possibility}\:{is}\:{that}\:{both}\:{have} \\ $$$${the}\:{last}\:{digit}\:\mathrm{1}.\:{that}\:{means}\:\mathrm{7}^{{k}} =\left(\mathrm{7}^{\mathrm{2}} ×\mathrm{7}^{\mathrm{2}} \right)^{{n}} , \\ $$$${i}.{e}.\:{k}=\mathrm{4}{n}.\:{we}\:{can}\:{prove}\:{that}\:\mathrm{7}^{\mathrm{4}{n}} −\mathrm{1} \\ $$$${is}\:{indeed}\:{a}\:{multiple}\:{of}\:\mathrm{15}. \\ $$

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

sir how do you get thier last digits please

$${sir}\:{how}\:{do}\:{you}\:{get}\:{thier}\:{last} \\ $$$${digits}\:{please} \\ $$

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

7 ⇒7 7×7 ⇒9 7×7×7 ⇒3 7×7×7×7 ⇒1 7×7×7×7×7 ⇒ 7 etc. 15×1+1 ⇒6 15×2+1 ⇒1 15×3+1 ⇒6 etc.

$$\mathrm{7}\:\Rightarrow\mathrm{7} \\ $$$$\mathrm{7}×\mathrm{7}\:\Rightarrow\mathrm{9} \\ $$$$\mathrm{7}×\mathrm{7}×\mathrm{7}\:\Rightarrow\mathrm{3} \\ $$$$\mathrm{7}×\mathrm{7}×\mathrm{7}×\mathrm{7}\:\Rightarrow\mathrm{1} \\ $$$$\mathrm{7}×\mathrm{7}×\mathrm{7}×\mathrm{7}×\mathrm{7}\:\Rightarrow\:\mathrm{7}\:{etc}. \\ $$$$ \\ $$$$\mathrm{15}×\mathrm{1}+\mathrm{1}\:\Rightarrow\mathrm{6} \\ $$$$\mathrm{15}×\mathrm{2}+\mathrm{1}\:\Rightarrow\mathrm{1} \\ $$$$\mathrm{15}×\mathrm{3}+\mathrm{1}\:\Rightarrow\mathrm{6}\:{etc}. \\ $$

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

here the proof that 7^(4n) ≡1 mod (15) 7^(4n) =2401^n =(2400+1)^n =Σ_(r=0) ^n C_r ^n ×2400^r =1+Σ_(r=1) ^n C_r ^n ×2400^r =1+Σ_(r=1) ^n C_r ^n ×(160×15)^r ≡1 mod (15)

$${here}\:{the}\:{proof}\:{that}\:\mathrm{7}^{\mathrm{4}{n}} \equiv\mathrm{1}\:{mod}\:\left(\mathrm{15}\right) \\ $$$$\mathrm{7}^{\mathrm{4}{n}} =\mathrm{2401}^{{n}} =\left(\mathrm{2400}+\mathrm{1}\right)^{{n}} =\underset{{r}=\mathrm{0}} {\overset{{n}} {\sum}}{C}_{{r}} ^{{n}} ×\mathrm{2400}^{{r}} \\ $$$$=\mathrm{1}+\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}{C}_{{r}} ^{{n}} ×\mathrm{2400}^{{r}} \\ $$$$=\mathrm{1}+\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}{C}_{{r}} ^{{n}} ×\left(\mathrm{160}×\mathrm{15}\right)^{{r}} \\ $$$$\equiv\mathrm{1}\:{mod}\:\left(\mathrm{15}\right) \\ $$

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

thanks so much sir

$${thanks}\:{so}\:{much}\:{sir} \\ $$

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

see also Q80580

$${see}\:{also}\:{Q}\mathrm{80580} \\ $$

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