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Question 355666: A woman with a basket of eggs finds that if she removes the eggs from the basket either two, three, four, five or six at a time, there is always one egg left. However, if she removes the eggs seven at a time, there are no eggs left. If the basket holds at least 300 eggs, how many eggs are in the womans basket?
Answer by Edwin McCravy(20055)   (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 either two, three, four, five or six at a time, there is always one egg left.
Stop there for a minute.
To be a multiple of 2, 3, 4, 5, and 6 an integer must contain the factors
2*2*3*5 or 60. So in order for a positive integer to leave a remainder of 1
when divided by 2,3,4,5 or 6 it must be 1 more than a multiple of 60, or an
integer of the form 60p+1 where p is a positive integer. Now we read the
rest:
However, if she removes the eggs seven at a time, there are no eggs left.
So it must also be a multiple of 7, or 7q where q is a positive integer, so
we have the linear Diophantine equation
60p + 1 = 7q
Write all numbers in terms of their nearest multiple of the smallest
coefficient in absolute value, which is 7. So we write 60 as 63 - 3,
and leave 1 as it is because the nearest multiple of 7 to 1 is 0.
(63 - 3)p + 1 = 7q
63p - 3p + 1 = 7q
Divide through by 7
9p - 3p/7 + 1/7 = q
Isolate fractions:
9p - q = 3p/7 - 1/7
The right side is positive, and the left side is an integer,
so both sides equal a positive integer say A, so
9p - q = A and 3p/7 - 1/7 = A
Clear the second of fractions:
3p - 1 = 7A
Write all numbers in terms of their nearest multiple of the smallest
coefficient in absolute value, which is 3. So we write 7 as 6 + 1,
and leave 1 as it is because the nearest multiple of 3 to 1 is 0.
3p - 1 = (6 + 1)A
3p - 1 = 6A + A
Divide thru by 3
p - 1/3 = 2A + A/3
Isolate fractions:
p - 2A = 1/3 + A/3
The left side is an integer and the right side is positive,
so both sides equal a positive integer, say B
p - 2A = B and 1/3 + A/3 = B
Clearing the second equation of fractions,
1 + A = 3B
A = 3B - 1
Substituting in p - 2A = B
p - 2(3B - 1) = B
p - 6B + 2 = B
p = 7B - 2
So
60p + 1 = 60(7B - 2) + 1 = 420B - 120 + 1 = 420B - 119
Thus she can have 420B - 119 eggs in her basket, where B is any
positive integer.
For B = 1, she can have 301 eggs
For B = 2, she can have 721 eggs
For B = 3, she can have 1141 eggs
...
etc., etc., etc.
Edwin
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