Question:
A person's birthday on 21\(^{st}\) July was on Wednesday. On what day will the Christmas fall in the same year?
A person's birthday on 21\(^{st}\) July was on Wednesday. On what day will the Christmas fall in the same year?
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Instead of adding up every month between the start date and the target date, try jumping forward from the start date in steps of exactly 7 days, since the day of the week repeats every week. Find a nearby date that is a whole number of weeks after the start date, then count only the small number of days left to reach the exact target date.
Updated On: Aug 21, 2026
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The Correct Option is D
Approach Solution - 1
Step 1: Understanding the Question:
We are given that the 21st of July in a particular year was a Wednesday. We need to find the day of the week for Christmas, which is on the 25th of December of the same year. The question is noted to be potentially incorrect, so we must rely on our calculation.
Step 2: Key Formula or Approach:
The method is to count the total number of days between the two dates and then find the number of "odd days". An odd day is a day remaining after dividing the total number of days by 7. The day of the week for the final date will be the starting day plus the number of odd days.
Step 3: Detailed Explanation:
Let's count the number of days from 21st July to 25th December.
Days remaining in July = 31 - 21 = 10 days
Days in August = 31 days
Days in September = 30 days
Days in October = 31 days
Days in November = 30 days
Days in December = 25 days Total number of days = 10 + 31 + 30 + 31 + 30 + 25 = 157 days.
Now, we find the number of odd days by dividing the total days by 7:
\[ \frac{157}{7} = 22 \text{ weeks and } 3 \text{ days remainder} \] So, there are 3 odd days.
The day of the week for Christmas will be Wednesday + 3 odd days.
Wednesday + 1 day = Thursday
Wednesday + 2 days = Friday
Wednesday + 3 days = Saturday.
Step 4: Final Answer:
The calculated day for Christmas is Saturday.
We are given that the 21st of July in a particular year was a Wednesday. We need to find the day of the week for Christmas, which is on the 25th of December of the same year. The question is noted to be potentially incorrect, so we must rely on our calculation.
Step 2: Key Formula or Approach:
The method is to count the total number of days between the two dates and then find the number of "odd days". An odd day is a day remaining after dividing the total number of days by 7. The day of the week for the final date will be the starting day plus the number of odd days.
Step 3: Detailed Explanation:
Let's count the number of days from 21st July to 25th December.
Days remaining in July = 31 - 21 = 10 days
Days in August = 31 days
Days in September = 30 days
Days in October = 31 days
Days in November = 30 days
Days in December = 25 days Total number of days = 10 + 31 + 30 + 31 + 30 + 25 = 157 days.
Now, we find the number of odd days by dividing the total days by 7:
\[ \frac{157}{7} = 22 \text{ weeks and } 3 \text{ days remainder} \] So, there are 3 odd days.
The day of the week for Christmas will be Wednesday + 3 odd days.
Wednesday + 1 day = Thursday
Wednesday + 2 days = Friday
Wednesday + 3 days = Saturday.
Step 4: Final Answer:
The calculated day for Christmas is Saturday.
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Approach Solution -2
Concept:
Step 1: Note the given date and the target date
21 July is a Wednesday. The target date is 25 December of the same year.
Step 2: Move forward in jumps of 7 days from 21 July
Adding 7 days repeatedly keeps the day fixed at Wednesday, so 28 July, 4 Aug, 11 Aug, 18 Aug, 25 Aug, and so on up to 22 December are all Wednesdays. This can be checked in one shot: from 21 July to 22 December is $10 + 31 + 30 + 31 + 30 + 22 = 154$ days, and $154 = 22 \times 7$, a whole number of weeks, so 22 December is also a Wednesday.
Step 3: Count the small remainder to 25 December
25 December is only $25 - 22 = 3$ days after 22 December.
Wednesday plus 1 day = Thursday
Thursday plus 1 day = Friday
Friday plus 1 day = Saturday
Final Answer: Christmas (25 December) falls on Saturday.
- Any date that is a multiple of 7 days away from a known date falls on the exact same day of the week.
- This lets us jump forward from the given date in steps of 7 days to land on a date very close to the target date, so only a small remainder needs to be counted instead of the entire gap.
Step 1: Note the given date and the target date
21 July is a Wednesday. The target date is 25 December of the same year.
Step 2: Move forward in jumps of 7 days from 21 July
Adding 7 days repeatedly keeps the day fixed at Wednesday, so 28 July, 4 Aug, 11 Aug, 18 Aug, 25 Aug, and so on up to 22 December are all Wednesdays. This can be checked in one shot: from 21 July to 22 December is $10 + 31 + 30 + 31 + 30 + 22 = 154$ days, and $154 = 22 \times 7$, a whole number of weeks, so 22 December is also a Wednesday.
Step 3: Count the small remainder to 25 December
25 December is only $25 - 22 = 3$ days after 22 December.
Wednesday plus 1 day = Thursday
Thursday plus 1 day = Friday
Friday plus 1 day = Saturday
Final Answer: Christmas (25 December) falls on Saturday.
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