If 16 July 2001 is Monday, then what will be the day of the week on 23 December 2006?

The correct option is Saturday.

To find the day of the week on 23 December 2006, given that 16 July 2001 is a Monday, we need to calculate the total number of odd days between 16 July 2001 and 23 December 2006.

Step 1: Calculate odd days from 16 July 2001 to 16 July 2006

The number of full years between 16 July 2001 and 16 July 2006 is 5 years.

During these 5 years:
- Standard (ordinary) years have 365 days = 52 weeks + 1 odd day.
- Leap years have 366 days = 52 weeks + 2 odd days.

Let us check the leap years in this interval:
- The year 2004 is a leap year (includes February 29, 2004).

So, among the 5 years (2001 to 2006):
- Number of leap years = 1 (2004)
- Number of ordinary years = 4 (2002, 2003, 2005, 2006 up to July 16)

Total odd days from years:
$(1\times 2)+(4\times 1)=2+4=6\text{ odd days}$

Thus, on 16 July 2006, the day would be 6 days ahead of Monday, which is Sunday.

Step 2: Calculate odd days from 16 July 2006 to 23 December 2006

Now, we calculate the remaining days month by month in 2006:
- July: Remaining days in July = 31 - 16 = 15 days.
- August: 31 days
- September: 30 days
- October: 31 days
- November: 30 days
- December: 23 days

Total remaining days = 15 + 31 + 30 + 31 + 30 + 23 = 160 days.

Now, divide the total number of days by 7 to find the odd days:
$160÷7=22\text{ weeks with a remainder of }6\text{ days}$

So, there are 6 odd days from 16 July 2006 to 23 December 2006.

Step 3: Total odd days and final calculation

Total odd days from 16 July 2001 to 23 December 2006 = 6 + 6 = 12 odd days.

Dividing 12 by 7 to get the net odd days:
$12÷7=1\text{ week with a remainder of }5\text{ days}$

Adding 5 days to Monday:
Monday + 5 days = Saturday.

Therefore, the day of the week on 23 December 2006 will be Saturday.