Question 357851: A woman with a basket of eggs finds that if she removes
the eggs from the basket 3 or 5 at a time, there is always 1
egg left. However, if she removes the eggs 7 at a time,
there are no eggs left. If the basket holds up to 100 eggs,
how many eggs does she have? Explain your reasoning.
Found 2 solutions by ankor@dixie-net.com, Edwin McCravy:
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! A woman with a basket of eggs finds that if she removes
the eggs from the basket 3 or 5 at a time, there is always 1
egg left. However, if she removes the eggs 7 at a time,
there are no eggs left. If the basket holds up to 100 eggs,
:
We know that the no. of eggs is a multiple of 7
The units digit is 1 or 6, because of the 5
That leaves 21, 56, 91
the 3 tell us it has to be 91 eggs
Answer by Edwin McCravy(20064)  (Show Source):
You can put this solution on YOUR website! A woman with a basket of eggs finds that if she removes
the eggs from the basket 3 or 5 at a time, there is always 1
egg left. However, if she removes the eggs 7 at a time,
there are no eggs left. If the basket holds up to 100 eggs,
how many eggs does she have? Explain your reasoning.
Suppose she has N eggs
>>...the eggs from the basket 3 or 5 at a time, there is always 1
egg left...<<
That means that N is 1 more than a multiple of both 3 and 5, which means
that it is 1 more than a multiple of 15. So we can write N as 15p + 1, where k
is a positive integer.
>>...However, if she removes the eggs 7 at a time, there are no eggs left...<<
So we have to find a positive integer of the form 15p + 1 that is a multiple of 7.
Try p=1. 15p + 1 = 15(1) + 1 = 16, not a multiple of 7.
Try p=2. 15p + 1 = 15(2) + 1 = 31, not a multiple of 7.
Try p=3. 15p + 1 = 15(3) + 1 = 46, not a multiple of 7.
Try p=4. 15p + 1 = 15(4) + 1 = 61, not a multiple of 7.
Try p=5. 15p + 1 = 15(5) + 1 = 76, not a multiple of 7.
Try p=6. 15p + 1 = 15(6) + 1 = 91, which is a multiple of 7, so that's the answer.
91 eggs. You can also do it by solving the Diopantine equation
15p + 1 = 7q
Are you supposed to do it that way? That would depend on what
course of mathematics you are taking.
91 = 30*3+1 = 28*5+1 = 13*7
So
when she takes out 30 groups of 3 eggs each, she takes out 90 eggs, and has 1 egg left,
when she takes out 18 groups of 5 eggs each, she takes out 90 eggs, and has 1 egg left,
when she takes out 13 groups of 7 eggs each, she takes out all 91 eggs and has none left.
Edwin
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