# How many odd days are there in 1600 years ? I calculated it but I guess the result is wrong?

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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what is an odd day? –  Gamamal Aug 5 '12 at 13:41
@Khromonkey In a year there a 365 days. It means 52 weeks + 1 odd day –  Suhail Gupta Aug 5 '12 at 13:47
Is today an "odd day", for example? Why/why not? –  Henning Makholm Aug 5 '12 at 13:48
@HenningMakholm I guess I had the incorrect knowledge.I knew odd day as 52 weeks + 1 . –  Suhail Gupta Aug 5 '12 at 14:06
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. :-)] –  ShreevatsaR Aug 5 '12 at 14:33

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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97 every 400 years, you meant to say. –  Harald Hanche-Olsen Aug 5 '12 at 14:05
As 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. –  Ross Millikan Aug 5 '12 at 14:09

1 ordinary year = 1 odd day; 1 leap year= 2 odd days; 100 years= 76 ordinary years + 24 leap years; total odd days= 76x1 + 24x2 = 124 odd days= 17 weeks + 5 days; in 100 yrs, there are 5 odd days; in 200 yrs 5x2= 3 odd days; in 300 yrs 5x3= 1 odd day; in 400 yrs (5x4)+1= 0 odd day; in 800 yrs (5x8)+2= 0 odd day; in 1200 yrs (5x12)+3=0 odd days; in 1600 yrs (5x16)+4= 0 odd days; and so on..........

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A span of 100 years may have 24 or 25 leap years. You do account for this when you get to 400, but it is not very clear. –  Ross Millikan Aug 6 '12 at 13:52

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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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.

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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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