Word Problems/math IMP

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felipe wrote at 2006-09-10 03:37:20
Hi Samantha.

Hi Steve.



This is the way I see it.



Since the number of eggs divided by 5 will produce a reminder of 1, then the # of eggs must end on either 1 or 6. (Like 26 /5 = 5 and R=1 or 61/5 = 20 and R=1.)

Also, the # canít be divisible by 2, which imply that the # can not end with a 6.

Now we know for sure that the number of eggs must end with a 1.

This reduces our list to all the multiples of seven that end with a 1, like:

21, 91, 161, 231, 301, 371, etc.

Now, letís check which of these numbers will be the first solution (Because it can be more than one solution?)

21 is out because it can be divided by 3

91 is out because 91/4 = 22 and R=3

161 is out because (Can you figure out why?) 161/3 =53 and R=2

231 out

And 301 is your first answer!

Samantha, can you find one or two more possible solutions?



Felipe




EL Maestro wrote at 2006-10-29 07:04:30
Since the number of eggs divided by 5 will produce a reminder of 1, then the # of eggs must end on either 1 or 6. (Like 26 /5 = 5 and R=1 or 61/5 = 20 and R=1.)

Also, the # canít be divisible by 2, which imply that the # can not end with a 6.

Now we know for sure that the number of eggs must end with a 1.

This reduces our list to all the multiples of seven that end with a 1, like:

21, 91, 161, 231, 301, 371, etc.

Now, letís check which of these numbers will be the first solution (Because it can be more than one solution?)

21 is out because it can be divided by 3

91 is out because 91/4 = 22 and R=3

161 is out because (Can you figure out why?) 161/3 =53 and R=2

231 out

And 301 is your first answer!

Samantha, can you find one or two more possible solutions?



I can see a pattern for this problem.

Every 420 you will get the next possible solution.

301 + 420 =721 a solution

721 + 420 = 1141 a solution.

Etc.



Felipe




NCP Student wrote at 2008-06-17 06:02:20
Hello Steve and Samantha,



I'm going to be an incoming freshman at Northside College Prep and as a summer program, I used the IMP book, which included this very problem, which my class broke up into groups and solved. I will simply copy and paste my answer that I gave in class.



ANSWERS: 301 eggs or 721 eggs.



Our solution was that this woman had 301 eggs with her. To get this answer, we went through a long process of trial and erroring many numbers until we finally hit 301 as the 43rd multiple of 7. My group and I knew that this was one of the correct answers because 301 is not divisible by 2, 3, 4, 5,  or 6, and is only divisible by 7, which is the reason why the woman always had one egg left over trying to put it into groups of 2, 3, 4, 5 or 6. There are other solutions, such as 721, as found in class. An easier way to get to an answer such as 721 would be to find a number that fit evenly into 2, 3, 4, 5, and 6, which would be just to multiply all of them together to get 720. Then, because the number was always one off and wasn't a multiple of any of those numbers, we simple add a one to reach 721 as an answer, not to forget that the answer we arrived to, 301, also works.



Basically, just multiply, 2, 3, 4, 5, and 6 to get a common number of 720, then add one so you can never group the eggs into those numbers without having one left over. 721 is divisible by 7, by the way.



I hope I've helped. I see this question is about 4 years old, haha.


gooogoooo wrote at 2009-09-03 01:02:35
There is an answer. The answer must end in 1 because: it has to be odd because of 2. It has to be 1 more than a multiple of 5 because of 5, and it can't end in 6 because 6 is an even number. So then you go through every 70 numbers to find one that ends in one and is one more than a multiple to every number (besides 7). So the ANSWER is 301.  


jborges3 wrote at 2013-01-11 04:22:35
This question does have a solution, the number of eggs divided by 2,3,4,5 or 6 must have a remainder of 1. To get this we can simply multiply 2*3*4*5*6 and then add 1. This gives us 721.



Now we can check that 721 divided by each of those numbers gives a remainder of 1 (but it must because of the way we came up with it)



721/2 = 3*4*5*6 R1

721/3 = 2*4*5*6 R1

721/4 = 2*3*5*6 R1

721/5 = 2*3*4*6 R1

721/6 = 2*3*4*5 R1



By our luck, 721 happens to be divisible by 7, if it wasn't we could have multiplied it by 7 a few times to get other working answers of 146461 or 1731121. The question should be what is the lowest number of eggs that meets these conditions.


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