Science:Math Exam Resources/Courses/MATH312/December 2005/Question 07

MATH312 December 2005

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Question 07

Multiple Choice Let n be the solution to the following ancient Indian problem (taken from Rosen) If eggs are removed from a basket 2, 3, 4, 5 and 6 at a time, there remain respectively, 1, 2, 3, 4 and 5 eggs. But if the eggs are removed 7 at a time, no eggs remain. What is the least number of eggs that could have been in the basket? The number n is congruent to which of the following modulo 13? (a) 1 (b) 2 (c) 3 (d) 4 (e) None of the above.

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
For the quick solution, consider what happens when you add one egg to the basket.

Hint 2
The slower way uses a more standard approach with the Chinese Remainder Theorem.

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 1

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. The answer is b = 2. The quickest solution is to note that adding one egg to the basket gives a multiple of 2,3,4,5 and 6 and is congruent to 1 modulo 7. Since $\displaystyle {\text{lcm}}(2,3,4,5,6)=60$ we are simply looking for multiples of 60 that are congruent to 1 modulo 7. Notice that $\displaystyle 60\equiv 4\mod {7}$ and $\displaystyle 120\equiv 1\mod {7}$. Hence the answer is $\displaystyle 120-1=119$. reducing modulo 13 gives $119\equiv 2\mod {13}$

Solution 2

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. The correct answer is b = 2 The slower, albeit more constructive way, to solve this is to set up a system of equations consisting of the following ${\begin{aligned}n&\equiv 1\mod 2\quad n\equiv 2\mod {3}\quad n\equiv 3\mod {4}\\n&\equiv 4\mod {5}\quad n\equiv 5\mod {6}\quad n\equiv 0\mod {7}\end{aligned}}$ Now we just crunch condition by condition. Notice that some of these equations are redundant. In fact the cases modulo 2 and modulo 6 can be reduced from the cases modulo 4 and from combining the modulo 2 and 3 cases respectively. Hence we need only consider the equations ${\begin{aligned}n\equiv 2\mod 3\quad n\equiv 3\mod 4\quad n\equiv 4\mod 5\quad n\equiv 0\mod 7\end{aligned}}$ For the remaining steps, let $\displaystyle k_{i}$ be integers for each $\displaystyle i=1..4$ as needed. The first equation tells us that ${\begin{aligned}n=2+3k_{1}\end{aligned}}$ Plugging into the second yields $\displaystyle 2+3k_{1}\equiv 3\mod 4$ which implies that $\displaystyle k_{1}\equiv 3\mod 4$. This says $\displaystyle k_{1}=3+4k_{2}$. Plugging into the equation above yields ${\begin{aligned}n=2+3k_{1}=2+3(3+4k_{2})=11+12k_{2}\end{aligned}}$ Next, into the third equation yields $\displaystyle 11+12k_{2}\equiv 4\mod 5$ which implies that $k_{2}\equiv 4\mod 5$. This says $\displaystyle k_{2}=4+5k_{3}$ . Hence, plugging into the equation above yields ${\begin{aligned}n=11+12k_{2}=11+12(4+5k_{3})=59+60k_{3}\end{aligned}}$ Lastly, into the fourth equation yields $\displaystyle 59+60k_{3}\equiv 0\mod 7$ which implies that $\displaystyle k_{3}\equiv 1\mod 7$. This says $\displaystyle k_{3}=1+7k_{4}$. Hence, plugging into the equation above yields ${\begin{aligned}n=59+60k_{3}=59+60(1+7k_{4})=119+420k_{4}\end{aligned}}$ Thus, in order for all these conditions to be satisfied, we need n to be congruent to 119 modulo 420. The minimum such number is 119. Modulo 13, we see that the answer is 2.

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