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

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.

Solution 1 

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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 we are simply looking for multiples of 60 that are congruent to 1 modulo 7. Notice that and . Hence the answer is . reducing modulo 13 gives 
Solution 2 

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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
For the remaining steps, let be integers for each as needed. The first equation tells us that
Plugging into the second yields which implies that . This says . Plugging into the equation above yields
Next, into the third equation yields which implies that . This says . Hence, plugging into the equation above yields
Lastly, into the fourth equation yields which implies that . This says . Hence, plugging into the equation above yields
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. 