Generally in all most all competitive examinations there are some questions on dates and days. Here we are giving an easy technique to find name of day on certain date.
| Points to remember: |
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1 |
calendar: Gregorian calendar |
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2 |
Number of days in Ordinary year: 365 |
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3 |
Number of days in Leap Year 366 |
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4 |
Number of days in the months of January, March, May, July, August, October and in December are 31. |
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5 |
Number of days in the months of April, June, September and November are 30. |
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6 |
Number of days in the month of February is 28 in ordinary year and 29 in Leap Year. |
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If a year is divided by 4 absolutely and year ended with 00 (ex 100, 200) is divided by 400 is called leap year. Years 100, 200,300, 500, 600, 700, 900, 1000, 1100, 1300, 1400, 1500, 1700, 1800, and 1900 are not the leap years. |
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Every year is having 52 weeks and one odd day in ordinary year and 2 odd days in leap year. (365/7 = 52 (weeks) and remainder = 1 (Day)). That we can under stand if to day (15-08-08) is Fri Day next year on this date the day will be Thursday. If the next year is a leap year the day difference will be two days. For example on 15-08-07 the day is Wednesday. |
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For every four hundred years the day difference will be 0. That means the same calendar will repeat. |
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Easy technique for finding day name: |
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Simply remember the following tables. |
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TABLE I |
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| Century |
Add Days |
Century |
Add Days |
16 |
8 |
20 |
8 |
17 |
6 |
21 |
6 |
18 |
4 |
22 |
4 |
19 |
2 |
23 |
2 |
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A same formula is used for remaining centuries. |
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TABLE II : Day Codes |
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| Day Code |
Day Name |
0 |
Mon Day |
1 |
Tuesday |
2 |
Wednesday |
3 |
Thursday |
4 |
Friday |
5 |
Saturday |
6 |
Sunday |
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TABLE III |
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| Non Leap Year |
Leap Year |
31/1 |
1/2 |
28/2 |
29/2 |
4/4 |
4/4 |
6/6 |
6/6 |
8/8 |
8/8 |
10/10 |
10/10 |
12/12 |
12/12 |
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Remember above three tables. |
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Examples: |
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| 1. |
Find day name of 22nd July, 1967? |
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| Step1 |
Take century part 19 from 1967: 19 |
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Take value from Table-I against 19: 2 |
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| Step2 |
Take the Year part of 1967: 67 |
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| Step3 |
Take the Year part of 1967: 67 |
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Divide 67 by 4 (Leap year interval) |
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67/4= 16 and reminder 3 |
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Ignore reminder |
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Take 16 |
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| Step4 |
Referral Date =Reminder part of (step1+Step2+Step3)/7 |
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(2+67+16)/7 |
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85/7 |
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Reminder =1. |
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Take the day name from Table II against 1 |
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Day name for 1 is Tuesday |
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1967 is not a leap year. |
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Take the referral dates from Table III |
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On 31/1, 28/2, 4/4, 6/6, 8/8, 10/10, 12/12 in 1967 the day will be Tuesday. |
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The day from question 22-07-1967 is between 6/6 and 8/8 |
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The difference between 22/7 and 6/6 is (22days in July and 24 days in June) = 46 |
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Divide 46 by 7 : Reminder =4 |
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Take reminder 4 |
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The require day will be 4th day from Tuesday |
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Count from Wed 1, Thu 2, Fri 3, Sat 4th day |
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The answer is Saturday |
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See another example: |
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Find the day name on 02-10-2008 |
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| Step1 |
For Century part 20: 8 |
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| Step2 |
For Year Part :08 |
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| Step3 |
08/4 =2 |
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| Step4 |
Reminder part of (Step1+Step2+Step3)/7 |
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= (8+8+2)/7 |
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=18/7 |
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=4 |
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Select day name from Table II for 4: Friday |
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2008 is a leap year |
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On 01/1, 29/2, 4/4, 6/6, 8/8, 10/10, 12/12 in 1967 the day will be Friday. |
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Difference between 02-10-08 and 10-10-08 is 8 days. |
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Reminder of (8/7) =1 |
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one day before Friday i.e. Thursday |
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02-10-08 is Thursday |
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Second alternative: |
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Difference between 08-08-08 and 02-10-08 is (23+30+2) 55 days
Reminder of (55/7) = 6 |
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Hence 6th day from Fri Day is Thursday |
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| Submitted by Siddhartha |
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