asking for a method very similar

sujoy

Junior Member
Joined
Apr 30, 2005
Messages
110
Here is a problem
this is not a problem from any book .It is my own difficulty.
I have solved a problem & I wonder if something very similar could be applied here as well .........i.e., here I give you two problems ; the first one is solved & if the the second one could also be solved in the same way is what I ask you .
prob1]] I have some money on monday morning,from which I spend half on food and from the remaining I spend$200 on charity[on a per day basis]. This continues till friday when I have only $25 left with me in the evening , after my spendings .How much did I have on wednesday?
soln]]let thie money I spend on each day be x , so on friday things may be defined as x-x/2-200=25 => x=450 => I spend $425 on friday morning
Thus on each day things may be defined as :
425*2^n-200 , for friday n=1, & for wednesday n=3 .....etc

now the prob I ask you is prob2]] :
In a sports contest medals were awarded over 6 days.On the 1st day 1 medal & 1/7th of the remaing medals were awarded. On the 2nd day 2 medals & 1/7 th of the remaining were awarded& so on till the 5th day.On 6th day all the remaining were awarded. what could be the minimum number of medals that were awarded during the 6 days
Also could it be solved by number system or modular algebra.............
I mean it is just a figment of my imagination......could it be?
regards
Sujoy
 
On the first, the "going backwards" equation is
y=2x+400
last day y=25
back 1 = 2*25+400=450
back 1 = 2*450+400=1300
back 1 = 2*1300+400=3000
etc.
The nth day equation needs more work

On the second,
the 6th day y=(6x/7-6)=0 for x=7
back 1 = 6x/7-5=7 for x=14
back 1 = 6x/7-4=14 for x=21
etc
 
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