Question 1020348: Special thanks to MATHCOUNTS alum, Brian Edwards, Ph.D., for
submitting the problems for this week’s Problem of the Week.
Tuesday’s date is 02/16/2016. Notice that the four digits
representing the month and day, 02/16, are the same four digits
in the year, 2016.
1)When written in the form MM/DD/YYYY, how many days in 2016
have the property that the four digits of the month and the day
can be rearranged to form the year?
2)What was the most recent year that did not have any days with
this property?
3)What was the last date in the 20th century that had this
property?
4)The number of days in 2016 that have this property is equal
to the number of days in 2061 that have this property. For what
pair of years 20XY and 20YX, where X and Y are distinct digits,
is this not the case?
Answer by Edwin McCravy(20054)   (Show Source):
You can put this solution on YOUR website! Special thanks to MATHCOUNTS alum, Brian Edwards, Ph.D., for
submitting the problems for this week’s Problem of the Week.
Tuesday’s date is 02/16/2016. Notice that the four digits
representing the month and day, 02/16, are the same four digits
in the year, 2016.
1)When written in the form MM/DD/YYYY, how many days in 2016
have the property that the four digits of the month and the day
can be rearranged to form the year?
MM can be 01,02,06,10,12
MM=01 leaves 26 possible for DD
MM=02 leaves 16 possible for DD
MM=06 leaves 12 and 21 possible for DD
MM=10 leaves 26 possible for DD
MM=12 leaves 06 possible for DD
Answer: 6 days
2)What was the most recent year that did not have any days with
this property?
1999, as no month can be 19, 91, or 99
3)What was the last date in the 20th century that had this
property?
The last one that had a 0, which was 1990, which had the date
09/19.
4)The number of days in 2016 that have this property is equal
to the number of days in 2061 that have this property. For what
pair of years 20XY and 20YX, where X and Y are distinct digits,
is this not the case?
It would have to be a case where MM/DD/20XY occurs yet
MM/DD/20YX does not occur, or vice-versa.
The only date of the calendar that does not occur every year is
Feb. 29.
So it has to be a case where 02/29/20XY and 02/29/YX do not
both occur. So that would be a case where 20XY and 20YX are
not both leap years. But X and Y have to be the digits 2 and 9.
Leap years are divisible by 4, and so 2092 will be a leap year,
but 2029 will not be.
Answer: 02/29/92 will occur, but 2029 will not be a leap year
and so "02/29/29" will not be a date at all.
The pair of years are 2029 and 2092
Edwin
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