1. Questions

MATH help!

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 (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.



301


30


CAN YALL TELL ME 60% OF WHAT NUMBER IS 40?

  1. 👍
  2. 👎
  3. 👁
  4. ℹ️
  5. 🚩
Tammi

Respond to this Question

First Name

Your Response

Similar Questions

  1. math

    a farmer examines 25 cartons of eggs 3 contain cracked eggs whats the best prediction that in the number of cracked eggs in the delivery of 500

  2. Physics

    A 3.0-kg cart is rolling along a frictionless, horizontal track towards a 1.2-kg cart that is held initially at rest. The carts are loaded with strong magnets that cause them to attract one another. Thus, the speed of each cart

  3. Physics Help

    A small, 200 g cart is moving at 1.50 m/s on an air track when it collides with a larger, 2.00 kg cart at rest. After the collision, the small cart recoils at 0.890 m/s. Q: What is the speed of the large cart after the collision?

  4. physics

    A 50. kg cart is moving across a frictionless floor at 2.0 m/s. A 70. kg child, riding in the cart, jumps off so that the child hits the floor with zero velocity. What was the velocity of the cart after the boy jumped?

  1. math

    a market woman bought 30 eggs for #700.if 4 eggs were broken and she sold the rest at #25 each.how much did she gain?

  2. physics

    A grocery shopper tosses a(n) 9.1 kg bag of rice into a stationary 17.6 kg grocery cart. The bag hits the cart with a horizontal speed of 7.0 m/s toward the front of the cart. What is the final speed of the cart and bag? Answer in

  3. mathematics,

    A farmer sells 7.6 kilograms of apples and pears at the farmer's market. 3/5 of this weight is apples, and the rest is pears. How many kilograms of pears did she sell at the farmer's market?

  4. math

    A farmer is using a random sample to predict the number of broken eggs in a shipment of 3,000 eggs. Using a calculator, the farmer generates the following random numbers. The numbers 1-250 represent broken eggs. 477, 2116, 1044,

  1. 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

  2. physics

    Cart A has a mass of 3 kg and is moving to the right at 1 m/s. Cart B has a mass of 2 kg and is at rest. The carts have a perfectly elastic collision. Cart B starts moving to the right at 1.2 m/s. What is the final velocity of

  3. physics

    A 14.0 kg cart is moving with a velocity of 7.25 m/s down a level hallway. A constant force of -10.0 N acts on the cart, and its velocity becomes 3.20 m/s. Assume that air resistance is negligible. (a) What is the change in

  4. Mathematics

    A market woman buys N eggs at the rate of 12eggs for 60.8naira.8 of the eggs wer broken,so she sold the remaining at the rate of 4 eggs for 24naira and make a profit of 12naira.How many eggs did she buy.

View more similar questions or ask a new question.