The 2nd of June is a Thursday in a certain year. Which day of the week is the 3rd of July in that year?

Question

Consider a spherical globe rotating about an axis passing through its poles. There are three points P, Q, and R situated respectively on the equator, the north pole, and midway between the equator and the north pole in the northern hemisphere. Let P, Q, and R move with speeds $v_p$, $v_q$, and $v_r$, respectively.
Which one of the following options is CORRECT?

Answer 1

This question can be solved by numbering the days of the week and using simple modular arithmetic instead of counting day by day.

Assign each weekday a number starting from Thursday as 0: Thursday = 0, Friday = 1, Saturday = 2, Sunday = 3, Monday = 4, Tuesday = 5, Wednesday = 6. Since 2nd June is Thursday, it corresponds to number 0.

Next, count the number of days from 2nd June to 3rd July. June has 30 days, so the 2nd of June to the 2nd of July spans 30 days, and one more day brings us to the 3rd of July, giving $30 + 1 = 31$ days in total.

To find the weekday number for the 3rd of July, take the remainder when 31 is divided by 7, since the weekday pattern repeats every 7 days: $31 = 4 \times 7 + 3$, so the remainder is 3.

Add this remainder to the starting weekday number: $0 + 3 = 3$. Looking back at the numbering, weekday number 3 corresponds to Sunday.

Thursday: this is weekday number 0, which would only be correct if the day gap were a multiple of 7, but 31 is not a multiple of 7.
Friday: this is weekday number 1, which would need a remainder of 1, not 3.
Saturday: this is weekday number 2, which would need a remainder of 2, not 3.
Sunday: this is weekday number 3, which matches the remainder found above exactly.

Let's summarize:

There are 31 days between 2nd June and 3rd July.
Dividing 31 by 7 leaves a remainder of 3, so the weekday shifts forward by 3 places.
Three places after Thursday is Sunday.

So the 3rd of July falls on a Sunday, which is option (D).

Answer 2

This question can be solved by numbering the days of the week and using simple modular arithmetic instead of counting day by day.

Assign each weekday a number starting from Thursday as 0: Thursday = 0, Friday = 1, Saturday = 2, Sunday = 3, Monday = 4, Tuesday = 5, Wednesday = 6. Since 2nd June is Thursday, it corresponds to number 0.

Next, count the number of days from 2nd June to 3rd July. June has 30 days, so the 2nd of June to the 2nd of July spans 30 days, and one more day brings us to the 3rd of July, giving $30 + 1 = 31$ days in total.

To find the weekday number for the 3rd of July, take the remainder when 31 is divided by 7, since the weekday pattern repeats every 7 days: $31 = 4 \times 7 + 3$, so the remainder is 3.

Add this remainder to the starting weekday number: $0 + 3 = 3$. Looking back at the numbering, weekday number 3 corresponds to Sunday.

Thursday: this is weekday number 0, which would only be correct if the day gap were a multiple of 7, but 31 is not a multiple of 7.
Friday: this is weekday number 1, which would need a remainder of 1, not 3.
Saturday: this is weekday number 2, which would need a remainder of 2, not 3.
Sunday: this is weekday number 3, which matches the remainder found above exactly.

Let's summarize:

There are 31 days between 2nd June and 3rd July.
Dividing 31 by 7 leaves a remainder of 3, so the weekday shifts forward by 3 places.
Three places after Thursday is Sunday.

So the 3rd of July falls on a Sunday, which is option (D).

The 2nd of June is a Thursday in a certain year. Which day of the week is the 3rd of July in that year?

Show Hint

The Correct Option is D

Solution and Explanation

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