Monday, April 13, 2009

An interesting problem...?

A class collects $50 to buy flowers for a classmate who is in the hospital. Roses cost $3 each and carnations cost $2 each. No other flowers are to be used. How many different bouquets could be purchased for exactly $50?





I%26#039;ve kind of attempted this problem without using any strategy.





25 $2s would make one 50. 48, 46 skip. but 44 would, meaning 22 $2 and 2 $3. and so i skip every two other %26quot;evens%26quot; and go all the way down to get the total different amounts.





but is there a less tedious way to solve it? thank you.

An interesting problem...?
Boil it down to an equation. Let c=number of carnations, and r=number of roses. Then we want





3r+2c=50, i.e. c=(50-3r)/2





Suffices to find number of integers r%26gt;=0 such that 50-3r is an even number%26gt;0. Easy to see that every other integer would not work, for example r=1 is not possible, so r must be an even number. Therefore r=0,2,4,6,...,16, since r=18 would give a negative number of carnations.





So 9 possible bouquets.





There is a more general theory behind this, we are only choosing r such that 3r is congruent to 50 mod 2, which is the same as wanting 2|3r. The only extra detail is ensuring that both r and c are always non-negative.
Reply:16 r + 1 c = $48 + $2


14 r + 4 c = $52 + $8


12 r + 7 c = $36 + $14


10 r + 10 c = $30 + $20


8 r + 13 c = $24 + $26


6 r + 16 c = $18 + $32


4 r + 19 c = $12 + $38


2 r + 22 c = $6 + $44


0 r + 25 c = $0 + $50





Answer: 9 ways. So far, I don%26#039;t know of any less tedious way.



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