Question 622939: I came across this question on the internet after reading about the recent Raytheon MATHCOUNTS competition. I guess there is a long way and a short way to solve the problem (the competition winner solved it in 19 seconds supposedly), and I was hoping you could show me both ways please. I am a very slow learner so I'd appreciate it if you don't skip too many steps when you show me the solution. Thank you in advance.
A bag of coins contains only pennies, nickels and dimes with at least five of each. How many different combined values are possible if five coins are selected at random?
Answer by richard1234(7193)   (Show Source):
You can put this solution on YOUR website! I actually competed at the National MATHCOUNTS competition, back in 2007 and 2008 when they were held in Ft. Worth and Denver (didn't come close to the countdown round though...).
Note that the number of pennies determines uniquely the value, modulo 5. Assume we have no pennies. Using nickels and dimes, we can make the values 25, 30, ..., 50 cents (six values).
If we have one penny, we can make 21, 26, ..., 41 cents (five values).
If we have two pennies, we can make 17, 22, 27, 32 cents (four values).
It turns out that as we add one more penny, the number of possible values decreases by 1 (because we have one fewer dime/nickel). Therefore the answer is 6+5+4+3+2+1 = 21.
|
|
|
