If 28th August 1946 was a Wednesday, what day of the week was 31 Augus...
It is given that 28th August 1946 was Wednesday.
From 28th August 1946 to 28th August 1961, we have 4 leap years and 11 normal years.
So the number of odd days would be 11*1 + 4*2 = 19
Now the date which is asked is 31 Aug 1961. So if we move from 28th August to 31st August, we will have 3 more odd days.
So total number of odd days = 5 + 3 = 8
Now 8 mod 7 = 1 .
So 31st August 1961 would be Wednesday + 1 = Thursday.
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If 28th August 1946 was a Wednesday, what day of the week was 31 Augus...
Explanation:
To determine the day of the week for August 31, 1961, we need to calculate the number of days between August 28, 1946, and August 31, 1961.
Step 1: Calculate the number of years between the two dates
Since we need to calculate the number of days, we will consider the leap years as well.
Number of leap years between 1946 and 1961 = (Number of leap years before 1961) - (Number of leap years before 1946)
To calculate the number of leap years before a given year, we use the formula:
Number of leap years = (year / 4) - (year / 100) + (year / 400)
Number of leap years before 1961 = (1961 / 4) - (1961 / 100) + (1961 / 400) = 490 - 19 + 4 = 475
Number of leap years before 1946 = (1946 / 4) - (1946 / 100) + (1946 / 400) = 486 - 19 + 4 = 471
Number of leap years between 1946 and 1961 = 475 - 471 = 4
Number of non-leap years between 1946 and 1961 = (1961 - 1946) - Number of leap years between 1946 and 1961 = 15 - 4 = 11
Step 2: Calculate the total number of days between the two dates
Number of days = (Number of leap years * 366) + (Number of non-leap years * 365) + (Number of days from August 28 to August 31)
Number of days = (4 * 366) + (11 * 365) + 3 = 1464 + 4015 + 3 = 5482
Step 3: Determine the day of the week
Since we know that August 28, 1946, was a Wednesday, we can calculate the day of the week for August 31, 1961, by finding the remainder when the total number of days is divided by 7.
5482 ÷ 7 = 826 remainder 4
Since August 28, 1946, was a Wednesday (represented by 0), adding the remainder 4 to 0 gives us a total of 4.
Conclusion:
Therefore, August 31, 1961, was a Thursday.