Wednesday
July 23, 2014

Homework Help: Math

Posted by Jessica on Friday, September 15, 2006 at 4:21pm.

A farmer was taking her eggs but they all fell on the floor. She didn't know how many eggs she had. But then she remembers when she put them in groups of two, there was one egg left over. When she put them in groups of three, there was also one egg left over. The same thing happened for the groups of four, groups of five , or groups of six. But when she put them in groups of seven she ended up with complete groups of seven with no egg left over.
what can the farmer figure out from this information about how many eggs she had?
Is there more than one possibility?


My friend, Hee-Jin, and I worked this out by trial and error.

We knew the number had to be a multiple of 7. The answer had to be an odd number (multiples of even numbers are even). Also the answer couldn't end in 5 (multiples of 5).

So -- we just kept multiplying 7 until Hee-Jin got to 7 x 343 = 2401. She multiplied 49 x 49 (multiples of 7) and found the answer! Please check to see if there's a smaller number that also answers this question.


BobPursley also points out that this site explains another way of finding the answer - by modulo math.

http://www.math.rutgers.edu/~erowland/modulararithmetic.html


This is an example of a famous result in number theory called the Chineese Remainder Theorem. The object is to determine a number from the remainders it gives to different divisors or moduli as they're called.
Let n be the nubmer of eggs the farmer had. For this problem we're told
2a+1=n
3b+1=n
4c+1=n
5d+1=n
6e+1=n
7f=n

Here's where we could make some observations. If the number has a remiander of 1 when divided by 4 then it will certainly have a remainder of 1 when divided by 2. Check that. The same is true about 3 and 6: if it has a remainder of 1 when divided by 6 it will leave 1 when divided by 3.
What we can conclude is that we need two factors of 2 and one factor of 3 in n-1. We also need on factor of 5 in n-1. Thus
n-1=4*3*5*k for some k and
n=4*3*5*k+1=60k+1
We now need 60k+1 divisible by 7 so 60k+1=7f=n
This is where we would use trial and error, without resorting to results from elementary number theory. Verify that the smallest k that satisfies the requirement is k=5 so n=301.
Now test it
2*150+1=301
3*100+1=301
4*75+1=301
5*60+1=301
6*50+1=301
7*43=301

Is there more than one possibility?
Yes, there are an infinite number of possibilities, of course this is in theory.
Verify that 60*12+1 will also satisfy the requirements of the problem. Generally, 60*(7p+5)+1=420p+301 where p is any positive integer will satisfy the requirements. To do this show that 2,3,4,5, and 6 divide 420p and will leave a remainder of 1 when 301 is divided by each. 7 divides both numbers.
I'll mention that the solution MsSue provided is for p=5.
To answer this question:
What can the farmer figure out from this information about how many eggs she had?
This is the most we should say about n without more information:
n=420p+301 where p is a non-negative integer. p=0 gives the smallest solution.



Version 1

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.




Another example:

<< There are fewer than 6 dozen eggs in a basket. If I count them by 2's, there is 1 left over. If I count them by 3's, there are 2 left over. Three are left if I count them by 4's. Four are left if I count by 5's. How many eggs
are there? >>

Letting the number of eggs be N, we can write:
1--N/2 = A + 1/2 or N = 2A + 1
2--N/3 = B + 2/3 or N = 3B + 2
3--N/4 = C + 3/4 or N = 4C + 3
4--N/5 = D + 4/5 or N = 5D + 4
5--Equating (1) and (2) we get 2A + 1 = 3B + 2 or 2A - 3B = 1 or A = (3B + 1)/2.
6--Only values of B that make (3B + 1) evenly divisible by 2 are valid.
7--By trial and error
A...2...5....8...11...14..17..20...23...26...29 (After defining the first two quantities, subsequent values all differ
B...1...3....5....7....9...11...13..15...17...19 by the same amount as the first two values)
8--From B = (4C + 1)/3 we get
9--B...3...7...11...15..19..23..27...31...35...39
C...2...5....8....11..14..17..20..23...26....29
10--From C = (5D + 1)/4 we get
11--C...4...9...14...19..24..29..34..39....44...49
D...3...7...11...15..19..23..27..31....35...39
12--From B = (5D + 2)/3 we get
13--B...4...9...14...19..24..29..34...39...44...49
D...2...5....8....11..14..17..20...23...26...29

By inspection, the set of A, B, C, and D's that is consistent throughout is highlighted in the table, A = 29, B = 19, C = 14, and D = 11. This results in N = 2(29) + 1 = 3(19) + 2 = 4(14) + 3 = 5(11) + 4 = 59.

Checking:
59/2 = 29 + 1
59/3 = 19 + 2
59/4 = 14 + 3
59/5 = 11 + 4


A slightly different approach that only avoids the initial trial and error:

Letting the number of eggs be N, we can write:
1--N/2 = A + 1/2 or N = 2A + 1
2--N/3 = B + 2/3 or N = 3B + 2
3--N/4 = C + 3/4 or N = 4C + 3
4--N/5 = D + 4/5 or N = 5D + 4
5--Equating (1) and (2) we get 2A + 1 = 3B + 2 or 2A - 3B = 1
6--Dividing through by 2 yields A - B - B/2 = 1/2
7--(B + 1)/2 must be an integer k making B = 2k - 1
8--Substituting back into (5) gives us 2A - 6k + 3 = 1 or A = 3k - 1
9--For k = 1 and up
k...1....2....3....4....5....6....7....8.... 9....10
A...2...5....8...11...14..17..20...23...26...29
B...1...3....5....7....9...11...13..15...17...19
10--Following the same procedure using 3B + 2 = 4C + 3, 4C + 3 = 5D + 4, and 3B + 2 = 5D + 4, we can derive an array of B and C, C and D, and B and D for values of k resulting in the following:
From 2A + 1 = 3B + 2
k...1....2....3....4....5....6....7....8.... 9....10
A...2...5....8...11...14..17..20...23...26...29
B...1...3....5....7....9...11..13...15...17...19
From 3B + 2 = 4C + 3
k...1....2....3....4....5....6....7....8.... 9....10
B...3...7...11...15..19..23..27...31...35...39
C...2...5....8....11..14..17..20..23...26....29
From 4C + 3 = 5D + 4
k...1....2....3....4....5....6....7....8.... 9....10
C...4...9...14...19..24..29..34..39....44...49
D...3...7...11...15..19..23..27..31....35...39
From 3B + 2 = 5D + 4
k...1....2....3....4....5....6....7....8.... 9....10
B...4...9...14...19..24..29..34...39...44...49
D...2...5....8....11..14..17..20...23...26...29

By inspection, the set of A, B, C, and D that is consistent throughout is highlighted in the table, A = 29, B = 19, C = 14, and D = 11. This results in N = 2(29) + 1 = 3(19) + 2 = 4(14) + 3 = 5(11) + 4 = 59.

Checking:
59/2 = 29 + 1
59/3 = 19 + 2
59/4 = 14 + 3
59/5 = 11 + 4

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