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Printjmc
prealgebra senior
Problem
Carolyn and Paul are playing a game starting with a list of the integers to The rules of the game are:
Carolyn always has the first turn.
Carolyn and Paul alternate turns.
On each of her turns, Carolyn must remove one number from the list such that this number has at least one positive divisor other than itself remaining in the list.
On each of his turns, Paul must remove from the list all of the positive divisors of the number that Carolyn has just removed.
If Carolyn cannot remove any more numbers, then Paul removes the rest of the numbers.
For example, if a possible sequence of moves is shown in this chart:
Note that Carolyn can't remove or on her second turn, and can't remove any number on her third turn.
In this example, the sum of the numbers removed by Carolyn is and the sum of the numbers removed by Paul is
Suppose that and Carolyn removes the integer on her first turn. Determine the sum of the numbers that Carolyn removes.
Carolyn always has the first turn.
Carolyn and Paul alternate turns.
On each of her turns, Carolyn must remove one number from the list such that this number has at least one positive divisor other than itself remaining in the list.
On each of his turns, Paul must remove from the list all of the positive divisors of the number that Carolyn has just removed.
If Carolyn cannot remove any more numbers, then Paul removes the rest of the numbers.
For example, if a possible sequence of moves is shown in this chart:
Note that Carolyn can't remove or on her second turn, and can't remove any number on her third turn.
In this example, the sum of the numbers removed by Carolyn is and the sum of the numbers removed by Paul is
Suppose that and Carolyn removes the integer on her first turn. Determine the sum of the numbers that Carolyn removes.
Solution
The list starts as
If Carolyn removes then Paul removes the remaining positive divisor of (that is, ) to leave the list Carolyn must remove a number from this list that has at least one positive divisor other than itself remaining. The only such number is so Carolyn removes and so Paul removes the remaining positive divisor of (that is, ), to leave the list Carolyn cannot remove either of the remaining numbers as neither has a positive divisor other than itself remaining.
Thus, Paul removes and
In summary, Carolyn removes and for a sum of and Paul removes and for a sum of
If Carolyn removes then Paul removes the remaining positive divisor of (that is, ) to leave the list Carolyn must remove a number from this list that has at least one positive divisor other than itself remaining. The only such number is so Carolyn removes and so Paul removes the remaining positive divisor of (that is, ), to leave the list Carolyn cannot remove either of the remaining numbers as neither has a positive divisor other than itself remaining.
Thus, Paul removes and
In summary, Carolyn removes and for a sum of and Paul removes and for a sum of
Final answer
8