CONCEPT OF “CALENDAR”-01

1. Question: What shall be the day on 1^st January, 2000 A D? [Ans. (d) Saturday]              (a) Wednesday (b) Sunday (c) Monday (d) Saturday (e) none of these

Solution: 2000- one leap yr+1

 = 0-2+1 = -1 ≡ +6 ≡ Saturday. Ans.

2. Question: What was the day on 31^st December, 1800 A D? [Ans. (a) Wednesday]            (a) Wednesday (b) Sunday (c) Monday (d) Friday (e) none of these

Solution: 1600+200

= 0+3 = 3 ≡ Wednesday. Ans.

3. Question: The first Republic day of India was celebrated on 26^th January, 1950. What was the day of the week on that day?  [Ans. (a) Thursday]

(a) Thursday (b) Sunday (c) Monday (d) Tuesday   (e) none of these

Solution: [1600+300+49+26] years

= 0+1+0+12+5 ≡ 4 ≡ Thursday. Ans.

Concept:

A.      Every year, which is divisible by 4 is a leap year; except centurieth year, which should divisible by 400 to be a leap year.

B.      The period of a particular day in a week is 7. So, a week contains zero odd days.

C.      A month of 30 days contains 2 odd days; whereas the month of 31 days contains 3 odd days.

D.      A simple year contains one odd day, whereas a leap year contains two odd days.

E.      A period of 100 years contain 5 odd days.

F.       A period of 200 years contain 3 odd days.

G.      A period of 300 years contain 1 odd day.

H.      A period of 400 years contain 0 odd days.

I.        1^st January, 1^st A D was Sunday.

J.       If counting starts from 1^st January, 1^st A D, then

K.      “0” stands for Sunday

L.      “1” stands for Monday

M.     “2” stands for Tuesday

N.      “3” stands for Wednesday

O.      “4” stands for Thursday

P.       “5” stands for Friday

Q.      “6” stands for Saturday

R.      “0” stands for Sunday

Detailed Explanation:

1. Solution: 1^st January, 2000 A D

A period of 400 years contains “0” odd days, so 5 times of 400, that is, a period of 2000 years will also contain “0” odd days.

Since 2000^th year is a leap year, so 1999 years contains 0-2 odd days. Now one day ahead is 1^st January, 2000. It will be

0-2+1 = -1 ≡ +6 ≡ Saturday. Ans.

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2. Solution: 31^st December, 1800 A D

A period of 400 years contains “0” odd days, so four times of 400 years, that is, 1600 years will also contain “0” odd days. Next 200 years will contain 3 odd days.

Therefore, a complete period of 1800 years will contain 0+3 = 3 odd days which corresponds to “Wednesday”- as the last day of 1800 years, that is, 31^st December, 1800 A D. Ans.

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3. Solution: 26^th January, 1950.

A period of 400 years contains “0” odd days, so four times of 400 years, that is, 1600 years will also contain “0” odd days. Next 300 years will contain 1 odd day.

Therefore, a period of 1900 years will contain 0+3 = 3 odd days.

Next 49 years are also completed, including 12 leap years. They contain: 49+12≡0+5≡5 odd days. [When 49 is divided by 7, the remainder is 0 and when 12 is divided by 7, the remainder is 5; as the period of a particular day of a week is 7]

For 26^th January, 26 more days are added, but 26≡5. [When 26 is divided by 7, the remainder is 5, as the period of a particular day of a week is 7]

Finally the whole process can be summarized as

[1600+300+49+26] years ≡  0+1+0+5+5 ≡ 11 ≡ 4 ≡ Thursday. Ans.

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