SOLUTION: A woman with a basket of eggs finds that if she removes the eggs from a basket 3 or 7 at a time there is always one egg left. However if she removes the eggs 5 at a time there are
Algebra ->
Customizable Word Problem Solvers
-> Misc
-> SOLUTION: A woman with a basket of eggs finds that if she removes the eggs from a basket 3 or 7 at a time there is always one egg left. However if she removes the eggs 5 at a time there are
Log On
Question 1161162: A woman with a basket of eggs finds that if she removes the eggs from a basket 3 or 7 at a time there is always one egg left. However if she removes the eggs 5 at a time there are no eggs left. If the basket holds up to 100 eggs how many eggs does she have? Found 2 solutions by Boreal, ankor@dixie-net.com:Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website! The second digit is either a 0 or a 5
Therefore, need a multiplier of 3 and 7 that ends in 9 or 4. There are no suitable ones for 9 that are evenly divisible by both 3 and 7
for 4, there is the product (3*7)*4 or 84
Then there would be 85 eggs, which satisfies the second part. ANSWER
Removing them 3 at a time, 84 could be removed
same for 7 at a time. No other product of 3 and 7 ends in four for both when < 100
You can put this solution on YOUR website! A woman with a basket of eggs finds that if she removes the eggs from a basket 3 or 7 at a time there is always one egg left
:
However if she removes the eggs 5 at a time there are no eggs left.
the number of eggs has to be a multiple of 5
:
If the basket holds up to 100 eggs how many eggs does she have?
Find a common multiple of 3 and 7 which is one less than a multiple of 5
that would be 84, so 85 eggs would satisfy this requirement
