The New Sudoku Players' Forum

by **udosuk** » Sun Apr 23, 2006 8:36 am

Depends... From 1990 to 1999 (inclusive), there are a total of 120 months.
So far in this millenium there have been 75 months (not counting the current one)...

It's like this trick... Stick up 3 fingers and wave in front of somebody, asking "how many fingers are there?"... If he answers 3 then you can say the answer is 5, if he says 5 you can say the answer is 10 (both hands), if he tries 10 you can say 20 (both hands of both persons)... The other person just can't win...

by **MCC** » Sun Apr 23, 2006 8:40 am

Ok I should have qualified it.

How many months in a year have 28 days :?:

MCC

by **Ruud** » Sun Apr 23, 2006 10:46 am

MCC wrote:How many months in a year have 28 days?

12 (some have a few more, like July)

by **Smythe Dakota** » Sun Apr 23, 2006 1:14 pm

If all months had just 28 days, there would be 13 months per year. That would add up to 364 days. The 365th day could be an "extra" holiday, with no day-of-the-week designation. In leap years there could be two such days.

That way, the same date (e.g. July 15) would be on the same day of the week every year. In fact, the 15th of EVERY month would be on the same day of the week, month to month, year to year.

That would end the current hassle of figuring out different vacation schedules every year. I think I'll propose this to the Pope. He could call the 13th month Benedictember.

Bill Smythe

by **MCC** » Sun Apr 23, 2006 3:29 pm

New Years day would always fall on a Sunday. A day of rest after all that indulgence on New Years eve.

Think about those who have to work that day.

Double time for working on a Sunday, at Bank holiday rates, brilliant.

MCC

by **tarek** » Sun Apr 23, 2006 9:40 pm

How about this....

Code: Select all: At what age would you [you means everybody] celebrate your birthday on the same day of the week as the day you were born ?

There are 3 correct answers.

tarek

by **Ruud** » Sun Apr 23, 2006 11:47 pm

tarek wrote:At what age would you [you means everybody] celebrate your birthday on the same day of the week as the day you were born ?

Now any individual will, upto the age of 28, have one of the following sets of same-day-of-week birthdays:

Code: Select all: 5,11,22,28 11,17,22,28 6,11,17,28 6,17,23,28

But for every person, only each 28th birthday is always on the same day of week.

So that gives us (assuming nobody lives to the age of 122):

Code: Select all: 28, 56 & 84

But can you guess my birthday?

Here is a clue: Solve the puzzle.

Code: Select all: . . .|. . .|7 4 9 8 . .|. . 4|1 . . . . 6|3 . .|. . . -----+-----+----- 6 . .|. 7 .|. . 4 . . .|. 3 6|. . . . 5 7|. . .|. . . -----+-----+----- . 4 .|. 5 .|. 2 . . . .|. . .|. 7 . 7 . 3|9 . .|8 . .

Ruud.

by **udosuk** » Mon Apr 24, 2006 5:19 am

Ruud, you commited the common mistake of ignoring the minority groups

, namely:

a) those born on 29th Feb (the unlucky ones who celebrate their birthdays once every 4 years)

b) those born on or before 28th Feb 1900 (don't tell me there're none of those living among us...)

For group (a), suppose they're born after 29th Feb 1896, their birthday would be on the same day of week as their born day when they're:
28,56,84,112,...

As for group (b), it gets quite complicated because as most of us know, 1900 was NOT a leap year.

According to this page, the current verified oldest living person was born in 1889, so I divide group (b) into the following subgroups, according to their birthdays:

b01) 1st March 1899 to 28th Feb 1900
b02) 1st March 1898 to 28th Feb 1899
b03) 1st March 1897 to 28th Feb 1898
b04) 1st March 1896 to 28th Feb 1897
b05) 29th Feb 1896
b06) 1st March 1895 to 28th Feb 1896
b07) 1st March 1894 to 28th Feb 1895
b08) 1st March 1893 to 28th Feb 1894
b09) 1st March 1892 to 28th Feb 1893
b10) 29th Feb 1892
b11) 1st March 1891 to 28th Feb 1892
b12) 1st March 1890 to 28th Feb 1891
b13) 1st March 1889 to 28th Feb 1890
b14) 1st March 1888 to 28th Feb 1889
b15) 29th Feb 1888

Here is the list of their ages in which the birthdays share the same day of week of their born dates:

b01) 6,12,17,23,34,40,45,51,62,68,73,79,90,96,101,107,118,124,...
b02) 6,12,23,29,34,40,51,57,62,68,79,85,90,96,107,113,118,124,...
b03) 12,18,23,29,40,46,51,57,68,74,79,85,96,102,107,113,124,...
b04) 7,12,18,29,35,40,46,57,63,68,74,85,91,96,102,107,113,124,...
b05) 32,60,88,116,...
b06) 6,12,17,23,34,40,45,51,62,68,73,79,90,96,101,107,118,124,...
b07) 6,12,23,29,34,40,51,57,62,68,79,85,90,96,107,113,118,124,...
b08) 6,12,18,23,29,40,46,51,57,68,74,79,85,96,102,107,113,124,...
b09) 6,12,18,29,35,40,46,57,63,68,74,85,91,96,102,107,113,124,...
b10) 32,60,88,116,...
b11) 5,7,12,17,23,34,40,45,51,62,68,73,79,90,96,101,107,118,124,...
b12) 12,23,29,34,40,51,57,62,68,79,85,90,96,107,113,118,124,...
b13) 6,12,18,23,29,40,46,51,57,68,74,79,85,96,102,107,113,124,...
b14) 6,12,18,29,35,40,46,57,63,68,74,85,91,96,102,107,113,124,...
b15) 32,60,88,116,...

Hence, except (b05), (b10) & (b15) (those who born on 29th Feb in the 19th century), the only common ages of the remaining 12 subgroups are:
12,40,68,96,124,...

As for your birthday, I tried to solve your "puzzle", but unfortunately found 3 different solutions (i.e. a "pseudoku puzzle"), and from the partly completed grid, I could only spot 2 valid dates:
476319582 (r3): 6th March 1958?
723941865 (r9): 9th April 1865? (making you older than what my source claims!

)

by **tarek** » Mon Apr 24, 2006 8:03 am

Indeed Ruud, it is 28

The Julian year has 365.25 days
1 week has seven days

4 years makes the number of days an integer....

The least common multiple of 4 & 7 is therefore 28

I initially said 4 correct answers, but as udosuk mentioned, before 1st of March 1900, the 112 would not hold....The same goes for people born after 28th Feb 1988....... (Because the Gregorian year is a bit shorter & therefore 1900 & 2100 are not leap years)

Therefore 28,56 & 84 would always be correct for everybody Currently living on this Earth

For your birthday, my guess is very close to udosuk's......The First of AUGUST 1958, which means Ruud that you passed your 28th birthday a good while ago

tarek

by **Ruud** » Mon Apr 24, 2006 8:23 am

udosuk wrote:I tried to solve your "puzzle", but unfortunately found 3 different solutions

Could that have been done on purpose?

6th March 1958

No

9th April 1865

No

First of AUGUST 1958

No

Ruud.

by **udosuk** » Mon Apr 24, 2006 9:17 am

tarek wrote:Therefore 28,56 & 84 would always be correct for everybody Currently living on this Earth

Not true. At least not for Maria Esther de Capovilla of Ecuador, whose list is:
6,12,18,23,29,40,46,51,57,68,74,79,85,96,102,107,113,124,...

I'm pretty sure someone who's reading here has a living great grandmother who has a list which does not contain any of 28,56,84...

I lost the partial solution to Ruud's pseudoku, but from my memory the questionable cells have candidates 1,5,8, so I guess his birthday is 15 Aug or 18 May...

by **udosuk** » Mon Apr 24, 2006 10:16 am

After resolving (partially) the pseudoku, one more try:

March 29, 1975 (the main diagonal reads "32975" following the only cell with 3 candidates...)

by **Ruud** » Mon Apr 24, 2006 10:50 am

So far, no good.

I will give one more hint:

My birthdate is clueless.

Ruud.

by **tarek** » Mon Apr 24, 2006 10:58 am

udosuk wrote:Not true. At least not for Maria Esther de Capovilla of Ecuador, whose list is:
6,12,18,23,29,40,46,51,57,68,74,79,85,96,102,107,113,124,...

That lady broke too many rules in her time...

I'm happy for Maria Esther de Capovilla of Ecuador or any of her contemporaries to prove my theories wrong

however she passed her 84th birthday long time ago, that is why I took off 112 & the possible 140.........

What I was after was that everyone living on this earth who are going to celebrate next their 28, 56 or 84th birthdays would celebrate it on the same day of the week as the day they were born (all of them..born 1922-2006) will not cross 1900 or 2100 on those birthdays....

In any case: Thanx for the info udosuk, I must compose this better next time...

tarek

by **Smythe Dakota** » Mon Apr 24, 2006 11:25 am

What barks, has four legs, wags its tail, and eats concrete?

Bill Smythe

The New Sudoku Players' Forum

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