Question and Answers Forum

All Questions      Topic List

Number Theory Questions

Previous in All Question      Next in All Question      

Previous in Number Theory      Next in Number Theory      

Question Number 137364 by liberty last updated on 02/Apr/21

Find the remainder 7^(30)  divide  by 10

$${Find}\:{the}\:{remainder}\:\mathrm{7}^{\mathrm{30}} \:{divide} \\ $$$${by}\:\mathrm{10}\: \\ $$

Answered by MJS_new last updated on 02/Apr/21

7^0 =1  7^1 =7  7^2 =49  7^3 =343  7^4 =2401  30=7×4+2 ⇒ answer is 9

$$\mathrm{7}^{\mathrm{0}} =\mathrm{1} \\ $$$$\mathrm{7}^{\mathrm{1}} =\mathrm{7} \\ $$$$\mathrm{7}^{\mathrm{2}} =\mathrm{49} \\ $$$$\mathrm{7}^{\mathrm{3}} =\mathrm{343} \\ $$$$\mathrm{7}^{\mathrm{4}} =\mathrm{2401} \\ $$$$\mathrm{30}=\mathrm{7}×\mathrm{4}+\mathrm{2}\:\Rightarrow\:\mathrm{answer}\:\mathrm{is}\:\mathrm{9} \\ $$

Answered by EDWIN88 last updated on 02/Apr/21

we have 7^(30)  = 49^(15)  ≡ 9^(15)  (mod 10)  we can apply Binomial Theorem    9^(15)  = (10−1)^(15)  = Σ_(k = 0) ^(15)  (((15)),((  k)) ) 10^(15−k) .(−1)^k    so clearly 9^(15)  ≡ −1 (mod 10) = 9 (mod 10)  therefore the remainder 7^(30)  : 10 is equal  to 9 .

$$\mathrm{we}\:\mathrm{have}\:\mathrm{7}^{\mathrm{30}} \:=\:\mathrm{49}^{\mathrm{15}} \:\equiv\:\mathrm{9}^{\mathrm{15}} \:\left(\mathrm{mod}\:\mathrm{10}\right) \\ $$$$\mathrm{we}\:\mathrm{can}\:\mathrm{apply}\:\mathrm{Binomial}\:\mathrm{Theorem} \\ $$$$ \:\mathrm{9}^{\mathrm{15}} \:=\:\left(\mathrm{10}−\mathrm{1}\right)^{\mathrm{15}} \:=\:\underset{\mathrm{k}\:=\:\mathrm{0}} {\overset{\mathrm{15}} {\sum}}\begin{pmatrix}{\mathrm{15}}\\{\:\:\mathrm{k}}\end{pmatrix}\:\mathrm{10}^{\mathrm{15}−\mathrm{k}} .\left(−\mathrm{1}\right)^{\mathrm{k}} \: \\ $$$$\mathrm{so}\:\mathrm{clearly}\:\mathrm{9}^{\mathrm{15}} \:\equiv\:−\mathrm{1}\:\left(\mathrm{mod}\:\mathrm{10}\right)\:=\:\mathrm{9}\:\left(\mathrm{mod}\:\mathrm{10}\right) \\ $$$$\mathrm{therefore}\:\mathrm{the}\:\mathrm{remainder}\:\mathrm{7}^{\mathrm{30}} \::\:\mathrm{10}\:\mathrm{is}\:\mathrm{equal} \\ $$$$\mathrm{to}\:\mathrm{9}\:. \\ $$

Answered by benjo_mathlover last updated on 02/Apr/21

By Euler phi Theorem   7^(ϕ(n))  = 1 (mod n)   we find ϕ(10) = 10(1−(1/2))(1−(1/5))=4  so 7^(30)  = (7^4 )^7 ×7^2  = 1×7^2  (mod 10)  = 49 (mod 10) = 9 (mod 10)  it follows that the remainder  is 9

$${By}\:{Euler}\:{phi}\:{Theorem}\: \\ $$$$\mathrm{7}^{\varphi\left({n}\right)} \:=\:\mathrm{1}\:\left({mod}\:{n}\right)\: \\ $$$${we}\:{find}\:\varphi\left(\mathrm{10}\right)\:=\:\mathrm{10}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\right)\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{5}}\right)=\mathrm{4} \\ $$$${so}\:\mathrm{7}^{\mathrm{30}} \:=\:\left(\mathrm{7}^{\mathrm{4}} \right)^{\mathrm{7}} ×\mathrm{7}^{\mathrm{2}} \:=\:\mathrm{1}×\mathrm{7}^{\mathrm{2}} \:\left({mod}\:\mathrm{10}\right) \\ $$$$=\:\mathrm{49}\:\left({mod}\:\mathrm{10}\right)\:=\:\mathrm{9}\:\left({mod}\:\mathrm{10}\right) \\ $$$${it}\:{follows}\:{that}\:{the}\:{remainder} \\ $$$${is}\:\mathrm{9}\: \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com