Question

when eggs in a basket are removed 2,3,4,5,6 at a time there remain, respectively, 1,2,3,4,5 eggs. when they are taken out 7 at a time, none are left over. find the smallest # of eggs that could have been contained in the basket.

Answer 1

Find the x-and y-intercept of
y=3/4x-3

Answer 2

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 (60x5 + 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.