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I had to calculate the number of odd days in 1600 years. I have read the answer to be equal to 0. But I don't get it to equal to 0.

This is the way I am calculating the number of odd days in 1600 years :

1600 years = 24 x 16 = 384 leap years   (100 years = 24 leap years) (because 100 years have 24 leap years) 1 leap year = 2 odd days (52 weeks + 2 odd days) 384 leap years = 384 x 2 = 768 odd days --(A) 1600 years = 1600 - 384 = 1216 ordinary years 1 ordinary year = 1 odd day (52 weeks + 1 odd day) 1216 ordinary years = 1216 x 1 = 1216 odd days--(B)  Total number of odd days = (A) + (B) = 768 + 1216 = 1984 odd days in 1600 years  and 1984 is not divisible by 7 ! 

Am I making a mistake ? If yes,what is it ?

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    My attempt at making sense is that an "odd day" is a day of the week that occurs an odd number of times. In a non-leap year with 52weeks+1day, six of the seven days of the week occur 52 times (an even number), while one of them occurs 53 times (an odd number). So the number of "odd days" in a year is the number of {Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday} that occur an odd number (53) of times. [I haven't thought about 1752 when the Gregorian change happened. :-)]2012-08-05

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Remember that a year divisible by $400$ is a leap year. Although $2100$ will not be a leap year. $2400$ will be.

So in $400$ years there are precisely $97$ leap years.

And yes, the calendar repeats every $400$ years, so the number of days in $1600$ years is divisible by $7$. For $400$ years, to the $(400)(364)$ days, just add $400+100-3$ (ordinary advance by $1$ day, plus 100 for the leap years sort of, minus $3$ for the multiples of $100$ that are not multiples of $400$). Then multiply by $4$. So for $1600$ we get $4(500-3)$ "additional" days. Your $1984$ was essentially computed correctly, except that we need $4$ additional days for the $4$ multiples of $400$.

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    $A$s this answer shows, there are 497 odd days in 400 years, which is divisible by 7, so the days of the week repeat every 400 years.2012-08-05
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While your definition of "odd days" is unclear to me, what I get is that if the number of days is a multiple of $7$, the number of odd days is $0$. In the following I'll therefore show that the number of days indeed is a multiple of $7$.

The rule for leap years is:

  • If the year is divisible by $400$, it is a leap year.
  • Otherwise, if it is divisible by $100$, it is not a leap year.
  • Otherwise, if it is divisible by $4$, it is a leap year.
  • Otherwise, it is not a leap year.

This means, the leap year rule has a $400$ year period. Therefore if we want to know how many days there are in $1600$ years, we can calculate the days in $400$ years, and multiply by $4$.

Now the number of days in $400$ years is $400\cdot 365 + \text{number of leap years}$. Now to calculate the number of leap years in $400$ years, we go through the above list in reverse order:

  • Years divisible by $4$ are generally leap years, there are $400/4=100$ of them.
  • However this way we have also counted the years divisible by $100$ (because they are all divisible by $4$). There are $400/100=4$ of them, which we therefore subtract, so we get $100-4=96$.
  • However now we have also removed the years divisible by $400$, of which there's one in $400$ years. Therefore we have to add that one back, so we end up with $97$ leap years in $400$ years.

Therefore $400$ years have altogether $146\,097$ days. Now it is not hard to check that this number indeed is a multiple of $7$. Therefore there are no odd days in $400$ years, and thus also not in $1600$.

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In $400$ years there would be $100$ leap years, but $3$ of them are suppressed. An ordinary year has one "odd day", and a leap year has an extra "odd day". Therefore there are $497$ odd days in $400$ years, which is divisible by $7$. A fortiori there remains no "odd day" in $1600$ years.

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Every 4th century is a leap year and no other century is a leap year. Therefore there will be 96 + 1=97 leap year and 303 ordinary years. So the equation becomes 303*1 +97*2 =497 odd days or 0 odd days.

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@Sahil u r making a mistake in finding no. of leap years in 1600 years.. here is my solution..

as every 100 yrs has 24 leap years. so

• first 300 yrs-24*3 leap yrs

• 4th 100 yrs-24+1 leap yrs because 400th year is also a leap yr as it is divisible by 400

• next 300 yrs-24*3 leap yrs

• 8th 100 yrs-25 leap yr

• next 300 yrs-24*3 leap yrs

• 12th 100 yr-25 leap yrs

• next 300 yrs-24*3 leap yrs

• 16th 100 yrs-25 leap yrs

so now total no. of leap yrs in first 1600 yrs becomes=24*3*4+25*4=388

so no. of non leap yrs=1600-388=1212

so total no. of odd days in 1600 yrs=388*2+1212=1988 which is divisible by 7. So there're 0 odd days in 1600 yrs.

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It is easy. Just remembr the answer. :) Just joking. Here is my simple solution: As each 4th century's a leap year & no other century is a leap year, there will be 96 + 1=97 leap year and 303 ordinary years. So we get 303x1 + 97x2 =497 odd days or 0 odd days. Hope it helped. Afterall, better late than never, huh?

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If the number is divisible by $400$ then zero odd days.

In $1200$ or $1600$ years, you will find $0$ odd days.

In $100$ years, you will get $5$ odd days.