[Guest]

Hi,

Need help on this question. On which of the two days of the week, Saturday or Sunday, does New Year's Day fall more often?

[Guest]

actually, both have the same probability... if the eve falls on saturday , then it wud be sunday next yr.. unless the following year is a leap yr,coz then it would be monday after saturday but this effect would be zero on the long run...

[Guest]

This is a coding test, which means computation is required.

[Guest]

Maybe you should limit the timeframe, because otherwise the same probability goes not only for Saturdays and Sundays, but any other day of the week

[Guest]

The days of the week will fall on a certain date 4 times in a 28 year cycle. With 7 days in a week, a leap year every 4 years, the cycle of days will repeat every 28 years.

This is known as the perpetual calendar.

[superprismatic]

The Gregorian calendar cycles through all possible

yearly calendars in 400 years. You can see this

because the number of days in 400 years is

(3*365+366)*100-3 which is 146,097, a multiple

of 7. In 400 years, New Years Day is on a

Sunday 58 times and on a Saturday 56 times.

So, the answer to the OP is Sunday.

[Guest]

I completely disagree with you and believe that the chances are equal? Your formula is flawed by only looking at 400 years. You should instead look at a multiple of 28 since the pattern will repeat every 28 years due to leap years (7 days a week * 4 years to a leap year). So the breakdown would be New Year's falling on:

SU

M

T

W

F

SA

SU

M

W

TH

F

SA

M

T

W

TH

SA

SU

M

T

TH

F

SA

SU

T

W

TH

F

Year 29 would repeat the pattern of year 1. If you count the days New Year will fall out on in a 28 year period, it is completely even between every day of the week. So if you were to look at 420 years instead of 400 the results should be completely even.

The Gregorian calendar cycles through all possible

yearly calendars in 400 years. You can see this

because the number of days in 400 years is

(3*365+366)*100-3 which is 146,097, a multiple

of 7. In 400 years, New Years Day is on a

Sunday 58 times and on a Saturday 56 times.

So, the answer to the OP is Sunday.

[superprismatic]

I completely disagree with you and believe that the chances are equal? Your formula is flawed by only looking at 400 years. You should instead look at a multiple of 28 since the pattern will repeat every 28 years due to leap years (7 days a week * 4 years to a leap year). So the breakdown would be New Year's falling on:

SU

M

T

W

F

SA

SU

M

W

TH

F

SA

M

T

W

TH

SA

SU

M

T

TH

F

SA

SU

T

W

TH

F

Year 29 would repeat the pattern of year 1. If you count the days New Year will fall out on in a 28 year period, it is completely even between every day of the week. So if you were to look at 420 years instead of 400 the results should be completely even.

The rule for leap year is: Y is a leap year if and only if Y is a multiple of 4 AND if it's a multiple of 100, then it must also be a multiple of 400. So, the year 2000 was a leap year, but the year 1900 was not. That 400 in the rule means that the leap year cycle is 400 years long.

Edited November 18, 2009 by superprismatic

[Guest]

The rule for leap year is: Y is a leap year if and only if Y is a multiple of 4 AND if it's a multiple of 100, then it must also be a multiple of 400. So, the year 2000 was a leap year, but the year 1900 was not. That 400 in the rule means that the leap year cycle is 400 years long.

New years day moves forward one day every year in a normal year.

Every 4th year one day misses its chance to be new years day.

Starting with Saturday as new years day in year 0 the first year to miss will be Wednesday in year 4, then Monday in year 8 and Saturday in year 12, Thursday in year 16, Tuesday in year 20, Sunday in year 24, Friday in year 28 and then we are back to Wednesday in year 32.

Thus the pattern repeats over 28 years (32 – 4)

Now let’s consider what happens in year 400 (when we miss a leap year);

L 393 394 395 L 397 398 399 400 401 402 403 L 405 406 407 L 409 410 411 L

Sat Su M Tu * Th F Sa Su M Tu W Th * Sa Su M Tu * Th F Sa Su * Tu

413 414 415 L 417 418 419 L

W Th F * Su M Tu W * F

(sorry folks, try as I might I cant get these to line up properly in the text window)

Clearly in the 28 year cycle containing the 400th year the whole pattern resets one day earlier in the week. What this means is that in our example every 28th year NYD will be a Saturday up to year 392 and then multiples of 28 after that will be Fridays from years 420 to 784 etc as shown below;

1 – 392 = Sat 400 reset a day earlier

420 – 784 = Fri 800 reset a day earlier

812 – 1176 = Th 1200 reset a day earlier

1204 - 1596 = Wed 1600 reset a day earlier

1624 to 1988 = Tues 2000 reset a day earlier

2016 to 2380 = Mon 2400 reset a day earlier

2408 to 2772 – Sun 2800 Reset a day earlier

2800 to 2828 - Sat

So that pattern repeats every 2800 years

In essence, the answer is that without a time frame the question cannot be answered

Steve

[superprismatic]

New years day moves forward one day every year in a normal year.

Every 4th year one day misses its chance to be new years day.

Starting with Saturday as new years day in year 0 the first year to miss will be Wednesday in year 4, then Monday in year 8 and Saturday in year 12, Thursday in year 16, Tuesday in year 20, Sunday in year 24, Friday in year 28 and then we are back to Wednesday in year 32.

Thus the pattern repeats over 28 years (32 – 4)

Now let’s consider what happens in year 400 (when we miss a leap year);

L 393 394 395 L 397 398 399 400 401 402 403 L 405 406 407 L 409 410 411 L

Sat Su M Tu * Th F Sa Su M Tu W Th * Sa Su M Tu * Th F Sa Su * Tu

413 414 415 L 417 418 419 L

W Th F * Su M Tu W * F

(sorry folks, try as I might I cant get these to line up properly in the text window)

Clearly in the 28 year cycle containing the 400th year the whole pattern resets one day earlier in the week. What this means is that in our example every 28th year NYD will be a Saturday up to year 392 and then multiples of 28 after that will be Fridays from years 420 to 784 etc as shown below;

1 – 392 = Sat 400 reset a day earlier

420 – 784 = Fri 800 reset a day earlier

812 – 1176 = Th 1200 reset a day earlier

1204 - 1596 = Wed 1600 reset a day earlier

1624 to 1988 = Tues 2000 reset a day earlier

2016 to 2380 = Mon 2400 reset a day earlier

2408 to 2772 – Sun 2800 Reset a day earlier

2800 to 2828 - Sat

So that pattern repeats every 2800 years

In essence, the answer is that without a time frame the question cannot be answered

Steve

You forgot to take into account that years 100, 200, and 300 are not leap years.

[superprismatic]

New years day moves forward one day every year in a normal year.

Every 4th year one day misses its chance to be new years day.

Starting with Saturday as new years day in year 0 the first year to miss will be Wednesday in year 4, then Monday in year 8 and Saturday in year 12, Thursday in year 16, Tuesday in year 20, Sunday in year 24, Friday in year 28 and then we are back to Wednesday in year 32.

Thus the pattern repeats over 28 years (32 – 4)

Now let’s consider what happens in year 400 (when we miss a leap year);

L 393 394 395 L 397 398 399 400 401 402 403 L 405 406 407 L 409 410 411 L

Sat Su M Tu * Th F Sa Su M Tu W Th * Sa Su M Tu * Th F Sa Su * Tu

413 414 415 L 417 418 419 L

W Th F * Su M Tu W * F

(sorry folks, try as I might I cant get these to line up properly in the text window)

Clearly in the 28 year cycle containing the 400th year the whole pattern resets one day earlier in the week. What this means is that in our example every 28th year NYD will be a Saturday up to year 392 and then multiples of 28 after that will be Fridays from years 420 to 784 etc as shown below;

1 – 392 = Sat 400 reset a day earlier

420 – 784 = Fri 800 reset a day earlier

812 – 1176 = Th 1200 reset a day earlier

1204 - 1596 = Wed 1600 reset a day earlier

1624 to 1988 = Tues 2000 reset a day earlier

2016 to 2380 = Mon 2400 reset a day earlier

2408 to 2772 – Sun 2800 Reset a day earlier

2800 to 2828 - Sat

So that pattern repeats every 2800 years

In essence, the answer is that without a time frame the question cannot be answered

Steve

But 400 is a leap year. (sorry, I forgot to put this in my previous post)

[superprismatic]

Let's start with the year 2000 when New Years was on a Saturday.

Let 0=Sunday, 1=Monday, 2=Tuesday, 3=Wednesday, 4=Thursday, 5=Friday,

and 6=Saturday. I have written the New Years day numbers on a width

of 28. The year 2000 is a leap year but 2100, 2200, and 2300 are not.

Other years divisible by 4 are leap years. Here's the table:

New Year day numbers Years Comments
6123460124560234501235601345 2000-2027 2000 is a leap year
6123460124560234501235601345 2028-2055
6123460124560234501235601345 2056-2083
6123460124560234560124560234 2084-2111 in 2100 it's a 5 (Friday) and not a leap year
5012356013456123460124560234 2112-2139
5012356013456123460124560234 2140-2167
5012356013456123460124560234 2168-2195
5012345602345012356013456123 2196-2223 in 2200 it's a 3 (Wednesday) and not a leap year
4601245602345012356013456123 2224-2251
4601245602345012356013456123 2252-2279
4601245602345012356012345012 2280-2307 in 2300 it's a 1 (Monday) and not a leap year
3560134561234601245602345012 2308-2335
3560134561234601245602345012 2336-2363
3560134561234601245602345012 2364-2391
35601345 2392-2399
612346..... 2400-etc. in 2400 it's a 6 (Saturday) and a leap year
and things are as they were in 2000, so another
400-year cycle begins.

If you count the occurance of each digit you find

0 (Sunday) occurs 58 times

1 (Monday) occurs 56 times

2 (Tuesday) occurs 58 times

3 (Wednesday) occurs 57 times

4 (Thursday) occurs 57 times

5 (Friday) occurs 58 times

6 (Saturday) occurs 56 times

[Guest]

It's a simple code (Gregorian calendar starts 15 October 1581, so the first January 1 is in 1582, and it's a Friday.)

Saturdays: 60

Sundays: 62

Run it for yourself

public class Leapyear

{

    // 0 = saturday, 1 = sunday, 2 = monday, 3 = tuesday, 4 = wednesday, 5 = thursday, 6 = friday

    public static void main()

    {

        int sat = 0, sun = 0;

        for (int year = 1582, day = 6; year <= 2009; year++) {

            if (day == 0) {sat++;}

            if (day == 1) {sun++;}

            if (isLeapYear(year)) {day += 2;}

            else {day++;}

            day %= 7;

        }

        System.out.println("Saturdays: " + sat + "\nSundays: " + sun);

    }


    public static boolean isLeapYear(int year)

    {

        boolean leap = false;        

        if (year % 4 == 0) {leap = true;}

        if (year % 100 == 0 && year % 400 != 0) {leap = false;}

        return leap;

    }


}


// Outputs:

// Saturdays: 60

// Sundays: 62

[superprismatic]

It's a simple code (Gregorian calendar starts 15 October 1581, so the first January 1 is in 1582, and it's a Friday.)

Saturdays: 60

Sundays: 62

Run it for yourself

{
// 0 = saturday, 1 = sunday, 2 = monday, 3 = tuesday, 4 = wednesday, 5 = thursday, 6 = friday
public static void main()
{
int sat = 0, sun = 0;
for (int year = 1582, day = 6; year <= 2009; year++) {
if (day == 0) {sat++;}
if (day == 1) {sun++;}
if (isLeapYear(year)) {day += 2;}
else {day++;}
day %= 7;
}
System.out.println("Saturdays: " + sat + "\nSundays: " + sun);
}

public static boolean isLeapYear(int year)
{
boolean leap = false;
if (year % 4 == 0) {leap = true;}
if (year % 100 == 0 && year % 400 != 0) {leap = false;}
return leap;
}

}

// Outputs:
// Saturdays: 60
// Sundays: 62

public class Leapyear

When your program is run on ANY 400 year period, say from 1582 to 1981, it always gets 58 Sundays and 56 Saturdays.

[Guest]

When your program is run on ANY 400 year period, say from 1582 to 1981, it always gets 58 Sundays and 56 Saturdays.

yes, it was meant for those who thought that the number of Saturdays and Sundays were the same

[superprismatic]

yes, it was meant for those who thought that the number of Saturdays and Sundays were the same

By the way, I couldn't compile your program. I got the following error:

jan1.c:1: error: expected unqualified-id before ‘public’

With a different C compiler, I got this error:

jan1.c:1: error: expected ‘=’, ‘,’, ‘;’, ‘asm’ or ‘__attribute__’ before ‘class’

I don't program in C or C++, so I don't know what the problem is.

Can you help?

[Guest]

By the way, I couldn't compile your program. I got the following error:

jan1.c:1: error: expected unqualified-id before ‘public’

With a different C compiler, I got this error:

jan1.c:1: error: expected ‘=’, ‘,’, ‘;’, ‘asm’ or ‘__attribute__’ before ‘class’

I don't program in C or C++, so I don't know what the problem is.

Can you help?

It's java

[Guest]

It's java

I don't know any of the C's

[superprismatic]

I don't know any of the C's

I tried Java compilation, I got the error:

----------

1. ERROR in jan1.java (at line 1)

public class Leapyear

^^^^^^^^

The public type Leapyear must be defined in its own file

----------

1 problem (1 error)

[Guest]

I tried Java compilation, I got the error:

----------

1. ERROR in jan1.java (at line 1)

public class Leapyear

^^^^^^^^

The public type Leapyear must be defined in its own file

----------

1 problem (1 error)

I use http://drjava.org/ for simple java stuff, just make a new class in it and copy the program and it should work.

It works for me

[Guest]

You forgot to take into account that years 100, 200, and 300 are not leap years.

Actually I made two mistakes. I thought that it you missed a leap year every 400 when in fact you miss it every 100 and keep it every 400 (sorry).

Nether the less I think the prociple still holds;

Amended version;

Clearly in the 28 year cycle containing the 100th year the whole pattern resets one day earlier in the week. What this means is that in our example every 28th year NYD will be a Saturday up to year 84 and then multiples of 28 after that will be Fridays from years 112 to 196 etc as shown below;

1 – 84 = Sat 100 reset a day earlier because missed leap year

112 – 196 = Fri 200 reset a day earlier because missed leap year

224 – 280 = Thurs 300 reset a day earlier because missed leap year

308 - 392 = Wed 400 NO EFFECT

420 to 476 = Wed 500 reset a day earlier because missed leap year

504 to 588 = Tues 600 reset a day earlier because missed leap year

616 to 696 – Mon 700 Reset a day earlier because missed leap year

724 to 780 - Sun 800 NO EFFECT

808 to 892 - Sun 900 reset a day earlier because missed leap year

920 to 976 - Sat 1000 reset a day earlier because missed leap year

Based on the above the pattern repeats every 900 years