10^{th} January 2105 will be a Wednesday. Wha | vedclass

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(A) Given: $10^{th}$ January $2105$ is a Wednesday.We need to find the day on $31^{st}$ December $1704$.First,calculate the number of years between $3

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Sunday

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(A) Given: $10^{th}$ January $2105$ is a Wednesday.We need to find the day on $31^{st}$ December $1704$.First,calculate the number of years between $31^{st}$ December $1704$ and $31^{st}$ December $2104$. This is $2104 - 1704 = 400$ years.$A$ period of $400$ years contains $0$ odd days (since $400$ is a multiple of $400$ and contains $97$ leap years).Now,calculate the days from $1^{st}$ January $2105$ to $10^{th}$ January $2105$,which is $10$ days.Total odd days $= 0 + 10 = 10$ days.$10 \pmod 7 = 3$ odd days.Since we are moving backward in time from $2105$ to $1704$,we subtract the odd days from the given day.Wednesday $- 3$ days $=$ Sunday.

$10^{th}$ January $2105$ will be a Wednesday. What was the day on $31^{st}$ December $1704$?

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