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?
So what is 2x3x4x5x6 + 1?
Does that work? Why?
I answered this question for Jessica a couple posts below. Post if you have questions on it.
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
ok i want to know the lcm of 68 and 84 but does this site have it nooo it doesn't this helped me 0/100
ok i want to know the lcm of 68 and 84 but does this site have it nooo it doesn't this helped me 0/100
To find the least common multiple (LCM) of two numbers, such as 68 and 84, you can follow these steps:
1. Find the prime factorization of each number:
- The prime factorization of 68 is 2^2 x 17.
- The prime factorization of 84 is 2^2 x 3 x 7.
2. Identify the highest power of each prime factor that appears in either number:
- The highest power of 2 is 2^2.
- The highest power of 3 is 3.
- The highest power of 7 is 7.
- The highest power of 17 is 17.
3. Multiply these highest powers together to find the LCM:
LCM = 2^2 x 3 x 7 x 17 = 1428.
Therefore, the LCM of 68 and 84 is 1428.
Regarding your frustration with not finding the LCM on this site, I apologize for the inconvenience. It seems that the site may not have provided the specific information you were looking for. However, I'm here to help and provide the explanation you need.