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Week 1 Problem C
Archis Thu Oct 09, 2008 8:30 am
It only happens once every 400 years. Although "leap days" occur once every four years, leap days in a year ending with two zeroes only happen once every four centuries. That's right-- there's a leap day in 2000, but there wasn't one in 1700, 1800 or 1900. What fraction of the years from 2001 through 4000 inclusive have leap days?
Archis- Admin
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Re: Week 1 Problem C
Steve Mon Oct 13, 2008 4:59 am
I almost had it... then i realized they happen every 4 years (i had started with 3) Oh well.. i'll try again.
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Re: Week 1 Problem C
Mira Holford Mon Oct 13, 2008 5:32 am
Steve wrote:I almost had it... then i realized they happen every 4 years (i had started with 3) Oh well.. i'll try again.
HAHA.............jk
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Re: Week 1 Problem C
Archis Mon Oct 13, 2008 5:47 am
My solution:
- Spoiler:
- Imagine every mutliple of 4 from 2001 to 4000 is a leap year. That would be from from 4*501 to 4*1000 which gives 500 assumed leap days.
Now every 400 year interval, as the problem states, 3 of the 4 years ending in "00" have no leap days. There are 2000 years in total between 4000 and 2001. That divided by 400 and multiplied by 3, leads 15 non-leap days ending in "00."
Oh dear! It seems as if we overcounted in our first calculation. The actual number of leap years would be 500-15 or 485.
Thus expressed as a common fraction of leap years to total years it is 485/2000 or 97/400 in simplest form.
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Re: Week 1 Problem C
Steve Mon Oct 13, 2008 6:56 am
Thats why i didn't post my solution. mine was REALLY differnt.
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Re: Week 1 Problem C
Mira Holford Mon Oct 13, 2008 8:46 am
Steve wrote:Thats why i didn't post my solution. mine was REALLY differnt.
o lol
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Steve- The Cool Cube
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Re: Week 1 Problem C
Mira Holford Tue Oct 14, 2008 12:48 am
Steve wrote:Archis how do you get all these
hes like a genius
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