If 5 May 2000 is a Friday, then what will be the day of the week on 12 August 2010? 

Step 1: Understand the concept of "odd days" A normal year has 365 days = 52 weeks + 1 odd day A leap year has 366 days = 52 weeks + 2 odd days Leap years are multiples of 4, except centuries (like 1900) must be divisible by 400. Step 2: Count total number of days between From 5 May 2000 to 12 August 2010 Break it into: From 5 May 2000 to 5 May 2010 → 10 years From 5 May 2010 to 12 August 2010 Step 3: Count odd days in 10 full years (2000–2010) Let’s count leap years and normal years between 2000 and 2010 (excluding 2010): Leap years: 2000, 2004, 2008 → 3 leap years Normal years: 2001, 2002, 2003, 2005, 2006, 2007, 2009 → 7 normal years Odd days: 3 leap years × 2 odd days = 6 odd days 7 normal years × 1 odd day = 7 odd days Total = 6 + 7 = 13 odd days = 13 mod 7 = 6 odd days Step 4: Count days from 5 May 2010 to 12 August 2010 Breakdown by month: May: 26 days (from 5th to 31st) June: 30 days July: 31 days August: 12 days
Total = 26 + 30 + 31 + 12 = 99 days Odd days in 99 days = 99 mod 7 = 1 odd day Step 5: Total odd days = 6 (from years) + 1 (from 2010) = 7 odd days = 0 odd days So, 12 August 2010 will be Friday – 1 = Thursday.