Science:Math Exam Resources/Courses/MATH312/December 2009/Question 03 (a)

MATH312 December 2009

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Question 03 (a)
Find the last three decimal digits of the number $\displaystyle 7^{999}$.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
It suffices to consider the number modulo 1000.

Hint 2
Euler's theorem will help.

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Since considering a number modulo 1000 gives the last 3 digits, it suffices to consider this number modulo 1000. Now $\displaystyle \phi (1000)=\phi (2^{3}\cdot 5^{3})=\phi (8)\phi (125)=(4)(100)=400$ Euler's Theorem (valid throughout since $\displaystyle \gcd(7,1000)=1$) states that $\displaystyle 7^{400}\equiv 7^{\phi (1000)}\equiv 1\mod {1000}$ and so we have $\displaystyle {\begin{aligned}7^{999}&\equiv 7^{400}\cdot 7^{400}\cdot 7^{199}\mod {1000}\\&\equiv 7^{199}\mod {1000}\end{aligned}}$ Let's examine $\displaystyle 7^{200}$. Notice that $\displaystyle \phi (8)=4\quad {\text{and}}\quad \phi (125)=100$ and so by Euler's Theorem, we have $\displaystyle 7^{200}\equiv (7^{4})^{50}\equiv 1\mod {8}$. Thus $\displaystyle 7^{200}=1+8s$ for some integer s. Further, $\displaystyle 1+8s\equiv 7^{200}\equiv (7^{100})^{2}\equiv 1\mod {125}$ and so $\displaystyle 8s\equiv 0\mod {125}$ giving $\displaystyle s=125t$ for some integer t. Thus, $\displaystyle 7^{200}=1+8s=1+1000t$. So $\displaystyle 7^{200}\equiv 1\mod {1000}$. Hence $\displaystyle 7^{199}\equiv 7^{-1}\mod {1000}$. To complete the proof, we need to find the inverse of 7 modulo 1000. We can do this via the Euclidean algorithm. $\displaystyle {\begin{aligned}1000&=7(142)+6\\7&=6(1)+1\end{aligned}}$ and back substituting gives $\displaystyle {\begin{aligned}1&=7-6(1)\\1&=7-(1000-7(142))(1)\\1&=7(143)-1000(1)\end{aligned}}$ and so the inverse of 7 modulo 1000 is 143. Hence $\displaystyle 7^{999}\equiv 7^{199}\equiv 7^{-1}\equiv 143\mod {1000}$ completing the question.

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