-One revolution of the Sun by the Earth requires 365 days, 5 hours, 48 minutes and 49.7 seconds. the 5 hours, 48 minutes and 49.7 second is approximated to 1/4 of a day. Every 4th year, one full day is added( leap year). How much of an error in minutes and seconds does this approximation produce every 4 years?( This time difference is corrected at some later date.)
-Find the result when the largest 4 digit number, all of whose digits are different, is subtracted from the smallest 5-digits number, all of whose digits are different .
first: what is the difference in 5h48'49.7" and 6hr?
answer: 11'10.3" check that. In four years, it is four times that.
second. smallest 5 digit number dealing with digits 0,1,2,3,4, and leading digit not zero: 10234
largest 4 digit number: 9876
5 hrs, 48 min, 49.7 sec
= 5/24 + 48/(24x60) + 49.7/(24x60x60)
= .2422442 of a day error
in 4 years that would be .9689676
so when we add a day we have a difference of
1 - .9689676 = .0310324 of a day off every 4 years
which is equal to 44 minutes and 41.2 seconds
BTW, the current rule is this:
If the year is divisible by 4, then you have a leap year UNLESS, the year was a century whose first 2 digits were not divisible by 4, then you did not have a leap year, otherwise the century would also be a leap year
Thus in 1900 we did not have a leap year
but in 2000 we did.
There is something called "Zeller's Congruence"
http://en.wikipedia.org/wiki/Zeller's_congruence
which uses the above rules to calculate what day of the week any particular date is. I used to assign my computer science class to make up a program than has as input the date of your birthday, and it was to tell you if that was a Monday, Tuesday, etc.
To calculate the error produced every 4 years due to the approximation of 1/4 of a day, we need to compute the difference between the actual time of one revolution of the Earth around the Sun and the time approximated using 365 days.
First, let's convert the extra time of 5 hours, 48 minutes, and 49.7 seconds into minutes.
5 hours is equivalent to 5 * 60 = 300 minutes.
48 minutes add 48 minutes = 348 minutes.
49.7 seconds can be approximated to 50 seconds, which is 50/60 = 0.8333 minutes.
The total extra time is 300 + 348 + 0.8333 = 648.8333 minutes.
Since this is the time for 4 years, we need to divide this by 4 to get the time difference every 4 years.
648.8333 / 4 = 162.2083 minutes.
Now, we convert the minutes into seconds by multiplying by 60.
162.2083 * 60 = 9732.5 seconds.
Therefore, the error produced every 4 years due to this approximation is approximately 162 minutes and 49.7 seconds (or 162 minutes and 50 seconds).
Regarding the second question, let's find the difference between the smallest 5-digit number, all of whose digits are different, and the largest 4-digit number, all of whose digits are different.
The smallest 5-digit number with different digits is 10234.
The largest 4-digit number with different digits is 9876.
To subtract the two numbers, we align them by place value:
9876
- 10234
Starting from the rightmost digit, subtract 4 from 6. This gives us 2.
Moving to the next digit, subtract 7 from 3. Since 3 is smaller than 7, we need to borrow from the next place value.
Borrowing 1 from the tens place makes the tens place value 2 and reduces the hundreds place value by 1. Now, we have:
10 (from borrowing)
9876
- 10234
Then, add 10 to 3, which gives us 13, then subtract 13 from 7 in the hundreds place. This results in 4.
Continuing with the next digit, we subtract 1 from 9, which gives us 8.
Finally, subtracting 2 from 1 is not possible, so we need to borrow again from the thousands place.
Borrowing 1 from the thousands place makes it 0, and adds 10 to the ones place. Now, we have:
100 (from borrowing)
9876
- 10234
Adding 10 to 0 gives us 10, then subtracting 10 from 1 gives us -9.
Therefore, the result of subtracting the largest 4-digit number, all of whose digits are different, from the smallest 5-digit number, all of whose digits are different, is -9310.