I - What day of week was July 4, 1776?
Step 1: The value for July is 6, so 6 plus 4 equals 10 and so 10 minus 7 (the largest multiple of 7 in 10) equals 3. Hold this 3 to Step 3.
Step 2: 76 minus 56 (the largest multiple of 28 in 76) equals 20; 20 divided by 4 equals an even 5; so, 20 plus 5 plus another 5 for the value for the 1700s from the Century table equals 30. We do not have to worry about subtracting 1 because the date is a Jan/Feb date, although note that it was a leap year because the division by 4 was even. Carry this 30 down to Step 3. (We could, but there is no need to waste time doing this, reduce this number by the largest multiple of 7 in it, which would make it 2 (30-28=2).
Step 3: The 3 from step 1 plus the 30 from step 2 is 33; 33 minus 28 (the highest multiple of 7 in 33) equals 5, and Thursday is the 5th day of the week.
Now let's find each of the Month, Day, and Year of July, 4, 1776, when that one element is missing and the Day of Week is known.
II - In what Month(s) in 1776 did the 4th fall on a Thursday?
Steps 1-3: From steps 1-3 we get 6 as follows: Step 1 gives us a 4 (? + 4). Step 2 gibes us a 2 (the reduced result, but without regard to any subtract of 1 for a leap year January/Feb month in the date), so we add the 4 plus 2 and we get 6.
Step 4:
We have a 6 from steps 1-3 and we need to get to a 5, a Thursday, so we need to add 7 to the 5 for 12, and 12 minus 6 equals 6.
Looking at the month table, the target months with the value of 6 were January (remember, it is 6 and not 0 because 1776 was a leap year), April and July. Of course, we know the answer is July.
III - What Day(s) in July 1776 fell on a Thursday?
Steps 1-3: Calculating step 1 of the basic formula using 0 for the Day we get a 6. (6 + ? =6). Calculating Step 2, we get a 2 (same as above), and since we know the month is July we do not have to worry about any adjustment of 1 for a January/February date in a leap year. So, 6 plus 2 equals 8, which equals 1 (taking out the excess 7s).
Step 4:
We have a value of 1 from steps 1-3 and we need to get to a value of 5, a Thursday, so 5 - 1 equals 4, the value of the Day we are looking for.
The first target Day of that month is the value from above, a 4, so the first target Day is the 4th.
Add 7, 14, 21 to the July 4th to get the 2nd, 3rd and 4th Thursdays and you get the July 13th, 20th and 27th.
IV - In what the Year or Years after (say) 1771 did July 4th fall on a Thursday?
Steps 1-3: Using 0 (?) for the Year, Steps 1-3 in the basic formula yield a result
of 3 as follows: From Step 1, 6 plus 4, or 10, which equals 3 after taking out the excess 7s. Step 2 is, of course, unknown so it is 0. So the total we have from Step 3 is 3.
Step 4:
We know we have a Thursday, which is a 5, and we have a value of 3 from above, so the value of the year we are looking for is 5 minus 3, which is 2.
If we start at 1771, the value for that year using the Step 3 formula is (71 – 56 = 15; 15/4 without the decimal is 3; 15 plus 3 plus 5 for the Century is 23; 23 minus 21 is 2, the value for 1771, but also the vale we are looking for. If we want to keep going, let's leap forward 4 years to 1775 and add 5 to our value, which makes the year value for 1775 a 7, which is a 0. We need to step forward one year to 1776 to find that the value for that year is a 1/2 since 1776 is a leap year and the first value relates to January and February because it is a leap year.
V - In what Century did July 4th fall on a Thursday in the year '76?
Steps 1-3: From Step 1 in the basic formula we get a 3 (6 + 4 = 10 – 7 = 3). In step 2, calculating the year without any Century value in it we get a 25 (76 – 56 = 20; [20/4] equals 5; 20 + 5 equals 25). So 25 plus 3 = 28 which equals 0.
Step 4:
To go from the 0 from steps 1-3 to the 5 we have for a Thursday, we need 5.
The 18th Century (the 1700s) have the value of 5, so the year was 1776.
