Science:Math Exam Resources/Courses/MATH312/December 2010/Question 03

MATH312 December 2010

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Question 03

Find all three-digit combinations that can occur as the last three digits of $\displaystyle a^{100}$ where a ranges over the integers. Give a simple criterion for predicting which of these possibilities occurs for a given a without computing $\displaystyle a^{100}$. Prove that your criterion is correct. Hint: Investigate $\displaystyle a^{100}$ modulo 8 and modulo 125 separately, then use the Chinese Remainder Theorem.

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 hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
The hint in the problem statement does give you a good starting point. Use Euler's Theorem to help with the reduction computations.

Checking a solution serves two purposes: helping you if, after having used the hint, 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. We proceed as in the hints. Modulo 8, we have that if a is even, then the residue of this number is 0 modulo 8. If it is odd, then Euler's Theorem tells us that since $\displaystyle \phi (8)=8(1-1/2)=4$, we have $a^{100}=(a^{4})^{25}=(a^{\phi (8)})^{25}\equiv 1^{25}\equiv 1\mod {8}$ Similarly, modulo 125 we have that if 5 divides a, then the residue is 0 and otherwise, Euler's Theorem tells us that since $\displaystyle \phi (125)=5^{3}(1-1/5)=100$, we have $a^{100}=a^{\phi (125)}\equiv 1\mod {125}$ This gives us four possible systems. Either ${\begin{aligned}a^{100}\equiv 0\mod {8}\quad &{\text{and}}\quad a^{100}\equiv 0\mod {125}\\a^{100}\equiv 1\mod {8}\quad &{\text{and}}\quad a^{100}\equiv 0\mod {125}\\a^{100}\equiv 0\mod {8}\quad &{\text{and}}\quad a^{100}\equiv 1\mod {125}\\a^{100}\equiv 1\mod {8}\quad &{\text{and}}\quad a^{100}\equiv 1\mod {125}\end{aligned}}$ As 8 and 125 are coprime, we have via the Chinese Remainder Theorem that there exists a unique solution to each of these system of equations modulo 1000 (the product of 8 and 125). We break this down into cases. Case 1: $\displaystyle a^{100}\equiv 0\mod {8}\quad {\text{and}}\quad a^{100}\equiv 0\mod {125}$. This case occurs when 10 divides a. In this case, by inspection, we can see that the last three digits here are given by 000 since 0 is the unique solution to this system. Case 2: $\displaystyle a^{100}\equiv 1\mod {8}\quad {\text{and}}\quad a^{100}\equiv 0\mod {125}$. This occurs when a is odd and divisible by 5. In this case, we have that $\displaystyle a^{100}=1+8s$ and so $\displaystyle 1+8s\equiv 0\mod {125}$. The inverse of 8 modulo 125 is given by the Euclidean algorithm: ${\begin{aligned}125&=8(15)+5\\8&=5(1)+3\\5&=3(1)+2\\3&=2(1)+1\end{aligned}}$ and back substituting gives ${\begin{aligned}1&=3-2(1)\\1&=3-(5-3(1))(1)\\1&=3(2)-5(1)\\1&=(8-5(1))(2)-5(1)\\1&=8(2)-5(3)\\1&=8(2)-(125-8(15))(3)\\1&=8(47)-125(3)\\\end{aligned}}$ so the inverse is 47. Thus ${\begin{aligned}8s&\equiv -1\mod {125}\\8(47)s&\equiv -47\mod {125}\\s&\equiv 125-47\mod {125}\\s&\equiv 78\mod {125}\end{aligned}}$ so $\displaystyle s=78+125t$. Thus recalling that $\displaystyle a^{100}=1+8s$ we have that $\displaystyle a^{100}=1+8(78+125t)=1+624+1000t=625+1000t$ and so the last 3 digits are 625 in this case. Case 3: $\displaystyle a^{100}\equiv 0\mod {8}\quad {\text{and}}\quad a^{100}\equiv 1\mod {125}$. This occurs when a is even and not divisible by 5. In this case, we have that $\displaystyle a^{100}=8s$ and so $\displaystyle 8s\equiv 1\mod {125}$. The inverse of 8 modulo 125 as computed above is 47. Thus $\displaystyle s\equiv 47\mod {125}$ and hence $\displaystyle s=47+125t$. Thus recalling that $\displaystyle a^{100}=8s$ we have that $\displaystyle a^{100}=8s=8(47+125t)=376+1000t$ and so the last 3 digits are 376 in this case. Case 4: $\displaystyle a^{100}\equiv 1\mod {8}\quad {\text{and}}\quad a^{100}\equiv 1\mod {125}$. This occurs when a is odd and not divisible by 5. In this case, we have that $\displaystyle a^{100}=1+8s$ and so $\displaystyle 1+8s\equiv 1\mod {125}$. The inverse of 8 modulo 125 as computed above is 47. Thus $\displaystyle s\equiv 0\mod {125}$ and hence $\displaystyle s=0+125t$. Thus recalling that $\displaystyle a^{100}=1+8s$ we have that $\displaystyle a^{100}=1+8s=1+8(0+125t)=1+1000t$ and so the last 3 digits are 001 in this case. This completes all cases.

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