SOLUTION: You have a total supply of $1000$ pieces of candy, and an empty vat. You also have a machine that can add exactly $10$ pieces of candy per scoop to the vat, and another machine th

Algebra ->  Divisibility and Prime Numbers -> SOLUTION: You have a total supply of $1000$ pieces of candy, and an empty vat. You also have a machine that can add exactly $10$ pieces of candy per scoop to the vat, and another machine th      Log On


   



Question 1207616: You have a total supply of $1000$ pieces of candy, and an empty vat. You also have a machine that can add exactly $10$ pieces of candy per scoop to the vat, and another machine that can remove exactly $3$ pieces of candy with a different scoop from the vat. When these two machines are done, there is only one piece of candy left in the vat. What is the smallest possible number of times the the first machine added candy to the vat?
Answer by ikleyn(52803) About Me  (Show Source):
You can put this solution on YOUR website!
.
You have a total supply of 1000 pieces of candy, and an empty vat. You also have a machine
that can add exactly 10 pieces of candy per scoop to the vat, and another machine that can remove
exactly 3 pieces of candy with a different scoop from the vat. When these two machines are done,
there is only one piece of candy left in the vat. What is the smallest possible number of times
the the first machine added candy to the vat?
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The answer is absolutely clear and obvious for everybody, who will read the problem
to the end.

        The smallest possible number of times
        the first machine added candy to the vat is 1 (one).         ANSWER


Explanation.  One is good and does work,  since if  10  pieces of candies were added sometime
using one scoop once,  then  1  candy was left when the other scoop took off  3  candies three times.


Evaluating the problem from the point of view of its contents,
I would say that it has very poor contents, where the word  "poor"
is used as an antonym to the word  "rich".


            Lord,  save from such problems in the future.