Problem:
Alex has 75 red tokens and 75 blue tokens. There is a booth where Alex can give two red tokens and receive in return a silver token and a blue token, and another booth where Alex can give three blue tokens and receive in return a silver token and a red token. Alex continues to exchange tokens until no more exchanges are possible. How many silver tokens will Alex have at the end?
Answer Choices:
A. 62
B. 82
C. 83
D. 102
E. 103
Solution:
After Alex makes m exchanges at the first booth and n exchanges at the second booth, Alex has 75−(2m−n) red tokens, 75−(3n−m) blue tokens, and m+n silver tokens. No more exchanges are possible when he has fewer than 2 red tokens and fewer than 3 blue tokens. Therefore no more exchanges are possible if and only if 2m−n≥74 and 3n−m≥73. Equality can be achieved when (m,n)=(59,44), and Alex will have 59+44=(E)103​ silver tokens.
Note that the following exchanges produce 103 silver tokens:
Exchange 75 blue tokensExchange 100 red tokensExchange 48 blue tokensExchange 16 red tokensExchange 9 blue tokensExchange 2 red tokens​Red Tokens100016031​Blue Tokens05021012​Silver Tokens25759199102103​​
The problems on this page are the property of the MAA's American Mathematics Competitions