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.
This is wrong. The answer is 119.
This is wrong. The answer shown above is for the scenario when removing 2, 3, 4, 5, 6 eggs at a time leads to a remainder of 1, NOT a remainder of 1, 2, 3, 4, 5 respectively. The actual answer is 119. (Use Chinese Remainder Theorem).
Find the x-and y-intercept of
y=3/4x-3
To solve this problem, let's work through it step by step.
Let's assume the total number of eggs in the basket is "x".
First, we know that when eggs are removed 2, 3, 4, 5, and 6 at a time, respectively, there remain 1, 2, 3, 4, and 5 eggs. This information can be represented by the following equations:
x ≡ 1 (mod 2)
x ≡ 2 (mod 3)
x ≡ 3 (mod 4)
x ≡ 4 (mod 5)
x ≡ 5 (mod 6)
We can solve this system of congruences to find a common solution.
Let's start with the first equation: x ≡ 1 (mod 2). This means x leaves a remainder of 1 when divided by 2. The smallest positive value of x that satisfies this congruence is 1 itself.
Now, substitute this value of x into the second equation: 1 ≡ 2 (mod 3). This equation is not satisfied, so we'll need to find the next multiple of 2 that satisfies the congruence. The next smallest positive value of x that satisfies this congruence is 1 + 2 = 3.
Continue this process for the remaining equations:
For the third equation: 3 ≡ 3 (mod 4), which is satisfied by x = 3.
For the fourth equation: 3 ≡ 4 (mod 5), which is not satisfied. The next smallest positive value of x that satisfies this congruence is 3 + 4 = 7.
For the fifth equation: 7 ≡ 5 (mod 6), which is not satisfied. The next smallest positive value of x that satisfies this congruence is 7 + 6 = 13.
We have now found a solution that satisfies all five congruences: x = 13. This means we have found a possible number of eggs in the basket, which is 13.
Therefore, the smallest number of eggs that could have been contained in the basket is 13.
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.