Tathagat has written several numbers that leave a remainder of 2 when divided by 6. He claims, 'If you add any three such numbers, the sum will always be a multiple of 6.' Is Tathagat's claim true?
Yes, Tathagat's claim is true.
Given: Numbers that leave remainder 2 when divided by 6 are of the form 6k + 2.
Calculation: (6a + 2) + (6b + 2) + (6c + 2) = 6(a + b + c) + 6 = 6(a + b + c + 1)
∴ The sum is always divisible by 6.
Explanation
The chapter discusses remainders and divisibility. Numbers leaving remainder 2 when divided by 6 can be expressed algebraically as 6k + 2 where k is any integer. When we add three such numbers, say (6a + 2), (6b + 2), and (6c + 2), the sum equals 6(a + b + c) + 6 = 6(a + b + c + 1). This sum clearly has 6 as a factor, proving it is always a multiple of 6. Example: 2 + 8 + 14 = 24 (divisible by 6), 14 + 20 + 26 = 60 (divisible by 6).
Solution Steps
Express numbers that leave remainder 2 when divided by 6 as 6k + 2.
Add three such numbers: (6a + 2) + (6b + 2) + (6c + 2).
Simplify to get 6(a + b + c + 1).
Conclude that the sum is always divisible by 6.