# math

posted by .

A peddler is taking eggs to the market to sell. The eggs are in a cart that holds up to 500 eggs. If the eggs are removed from the cart 2,3,4,5, or 6 at a time, one egg is always left over. If the eggs are removed 7 at time, no eggs are left over. How many eggs are in the cart?

• math -

First solve the simple problem of removing 2 and 3 eggs at a time, and there is nothing left over. The answer is therefore the LCM (Lowest Common Multiple) of 2 and 3, or 6.

If one egg is left over, we simply add one more to make 7 (or 1 to 12 to make 13, etc.)

Having done this, we are ready to find the number of eggs that has one left over when taken 2,3,4,5,6 at a time, namely the LCM of 2,3,4,5,6 (or its multiple) and add one.

This gives 60 as the LCM, or the multiples, such as 120, 180, 240, etc.
We will add one to these numbers to give 61, 121, 181, 241, etc. which will all satisfy the first five conditions (and less than 500).

However, the additional condition says that the number of eggs has to be divisible by 7. So by trying the numbers 61, 121, 181, 241, ... up to 421, we will find one that is divisible by 7, which is the answer required.

• math -

A more often quoted version of this problem is:

A farmer taking her eggs to the market hits a pothole. The eggs all wind up broken. She then goes to her insurance agent who tells the farmer how many eggs did you have? She doesn't remember but she does remember how she placed them. She knew for a fact that when she put them in groups of 2,3,4,5 and 6 she had one egg left over but when she put them in groups of 7, she had none left over. What is the minimum number of eggs that the farmer could have had?

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.

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

Equating 2A + 1 = 3B + 1, B = 2A/3
A...3...6...9...12...15...18...21...24...27...30...33...36...39...42...45...48...51...54...57...60...63...66...69...72...75
B...2...4...6....8....10...12...14...16...18...20...22...24...26...28...30...32...34...36...38...40...42...44...46...48...50
A...78...81...84...87...90...93...96...99...102...105...108...111...114...117...120...123...126...129...132...135...138
B...52...54...56...58...60...62...64...66....68.....70.....72....74.....76.....78.....80....82.....84.....86....88.....90.....92
A...141...144...147...150
B....94.....96.....98...100

Equating 3B + 1 = 4C + 1, C = 3B/4
B...4...8...12...16...20...24...28...32...36...40...44...48...52...56...60...64...68...72...76...80...84...88...92...96...100
C...3...6...9....12....15...18...21...24...27...30...33...36...39...42...45...48...51...54...57...60...63...66...69...72....75

Equating 4C + 1 = 5D + 1, D = 4C/5
C...5...10...15...20...25...30...35...40...45...50...55...60...65...70...75...80...85...90...95...100
D...4....8....12...16...20...24...28...32...36...40...44...48...52...56...60...64...68...72...76....8

Equating 5D + 1 = 6E + 1, E = 5D/6
D...6...12...18...24...30...36...42...48...54...60...64...68...72...76...80
E...5...10...15...20...25...30...35...40...45...50...55...60...65...70...75

Equating 6E + 1 = F, F = (6E + 1)/7
E...1...8...15...22...29...36...43...50...57...64...71...78...85...92...99
F...1...7...13...19...25...31...37...43...49...55...61...67...73...79...85

Equating F = 2A + 1, A = (7F - 1)/2
F..1...3....5....7...9...11...13...15...17...19...21...23...25...27...29...31...33...35....37....39....41....43...45....47...49
A..3..10..17..24..31..38...45...52...59...66...73...80...87...94..101.108.115..122..129..136..143..150..157..164..171

A tedious review of the data finds the highlighted set of numbers consistent throughout the data leading to
N = 2(150) + 1 = 3(100) + 1 = 4(75) + 1 = 5(60) + 1 = 6(50) + 1 = 7(43) = 301.

• math -

301 eggs

## Similar Questions

1. ### Math

I'm having alot of problems with the problem. I don't want the answer but I need to know how to figure it out to get the answer. To put it in another way I don't know where to start. A farmer takes to market in her cart, but she hit …
2. ### science

are these correct as to physical and chemical 1. Heat pan. [P] 2. Melt butter. [P] 3. Crack desired number of eggs into a mixing bowl. [P] 4 Add milk to eggs. [P] 5 Add salt and pepper. [P] 6 Beat eggs, milk and seasonings. [P] 7 Add …
3. ### college math

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.
4. ### algebra

a farmer is taking her eggs to market in her cart, but she hits a pothole, whick knocks over all the containers of eggs. Though she herself is unhurt, every egg is broken. So she goes to the insurance agent who asks her how many eggs …
5. ### algebra

A peddler is taking eggs to market to sell. The eggs are in a cart that holds up to 500 eggs. If the eggs are removed from the cart 2, 3, 4, 5,or 6 at a time, one egg is always left over. If the eggs are removed 7 at a time, no eggs …
6. ### math

6 eggs \$1.00 12 eggs \$2.00 18 eggs \$3.00 What is the constant proportionality that relates the price to the quantity of eggs?
7. ### Math

The number of eggs Meg collected over the week were: Monday=8 eggs, Tuesday =5 eggs, Wednesday=6 eggs , Thursday=7 eggs, Friday=6 eggs, Saturday=5 eggs. By Sunday the weeks average was 6 eggs per day. How many eggs were collected on …
8. ### Math Solving problem

In an egg and spoon race, children must transport eggs from one end of a course to the other. Points are awarded for each successfully transported, and points are deducted for each egg which drops and breaks. Jenny transported 5 eggs …
9. ### math

Alec bought 3 trays of eggs each containing 30 eggs. She paid sh270 for each tray of eggs. Nine eggs were eaten at home and the rest were sold. If Maria made a profit of 20%, how much did she sell each egg?
10. ### math

A farmer collected some eggs from his farm and found b eggs broken. He packed the remaining eggs in c egg cartons. Each egg carton can hold a dozen eggs and no eggs were leftover. Write an algebraic expression for the number of eggs …

More Similar Questions