a farmer is taking her eggs to market in her cart, but she hits a pothole, which knocks over all the containers of eggs
a farmer is taking her eggs to market in her cart, but she hits a pothole, which knocks over all the containers of eggs. Though she herself is unhurt, every egg is broken. She goes to her insurance agent, who asks her how many eggs she had. She says she doesn't know, but she remembers some things from various ways she tried packing the eggs. she knows that when she put the eggs into groups of two, there was one egg left over. When she put them into groups of three, there was also one egg left over. The same thing happened when she put them into groups of four, five, and isx. But she ended up with no eggs left over when she put them into groups of seven. How many eggs were there? is there more than one answer?
The simple approach
The L.C.M. (Lowest Common Multiple) of the numbers 2 through 6 inclusive is 2^2x3x5 = 60. The smallest number satisfying the divisors of 2 through 6 with remainders of 1 is therefore 60 + 1 = 61. Clearly, any multiple of 60 plus a 1 will satisfy these limited requirements. However, we are looking for a specific value of (60n + 1) that is evenly divisible by 7.or (60n + 1)/7. Dividing by 7, we get (60n + 1)/7 = 8n + 4n/7 + 1/7 or 8n + (4n + 1)/7 telling us that (4n + 1) must be a multiple of 7. Through observation, we can see that n = 5 is clearly the smallest integral value of n that will satisfy the condition. Therefore, the least number of eggs is (60×5 + 1) = 301. Checking: 301/2 = 150 + 1 301/3 = 100 + 1 301/4 = 75 + 1 301/5 = 60 + 1 301/6 = 50 + 1 301/7 = 43
If we were not interested in the minimum amount of eggs, you can logically ask the question, "What other values of n will produce other answers?" Well, very quickly, 12 and 19 work. N(n=12) = 60(12) + 1 = 721. Thus, 721/2 = 360 + 1, 721/3 = 240 + 1, 721/4 = 180 + 1, 721/5 = 144 + 1, 721/6 = 120 + 1, and 721/7 = 103. N(n=19) = 60(19) + ! = 1141. Do you see the pattern in the additional values of n, 5, 12, 19,…….? The soluton is rather straight forward when the remainders are constant. If the remainders are all different however, the solution takes on a quite different challenge and is most easily solved by means of the Chinese Remainder Theorem.
An algebraic approach evolves as follows: 1–We seek the smallest number N that meets the requirements specified above. 2–We already know that the number 61 satisfies all the divisions and remainders up through the divisor of 6. 3–What we now seek is N = 7A = 61 + 60n or 7A – 60n = 61 4–Dividing through by the smallest coefficient yields A – 8n – 4n/7 = 8 + 5/7 or (4n + 5)/7 = A – 8n – 8 5–(4n + 5)/7 must be an integer as does (8n + 10)/7 6–Dividing by 7 again yields n + n/7 + 1 + 3/7 7–(n + 3)/7 is also an integer k making n = 7k – 3. 8–For the lowest value of k = 1, n = 4 making N = 61 + 60(4) = 301.
Again, higher values of N are derivable by letting k = 2, 3, 4,…etc. For k = 2, n = 11 making N = 721 and k = 3 leads to n = 18 or N = 1141.
The lengthy approach
The solution of this type of problem can also be solved algebraically.
Letting N be the number of eggs being sought, we can write N/2 = A + 1/2 or N = 2A + 1 N/3 = B + 1/3 or N = 3B + 1 N/4 = C + 1/4 or N = 4C + 1 N/5 = D + 1/5 or N = 5D + 1 N/6 = E + 1/6 or N = 6E + 1 N/7 = F or N = 7F