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 .
Calculation: Sum of three such numbers:
∴ The sum is always divisible by 6.
Explanation
Numbers leaving remainder 2 when divided by 6 can be expressed algebraically as 6k + 2 for some integer k. When we add three such numbers, say 6a + 2, 6b + 2, and 6c + 2, we get 6(a + b + c) + 6 = 6(a + b + c + 1). This is clearly a multiple of 6, confirming Tathagat's claim.
Solution Steps
Numbers that leave remainder 2 when divided by 6 are of the form 6k + 2
Sum of three such numbers: (6a + 2) + (6b + 2) + (6c + 2)
Simplify: 6(a + b + c) + 6 = 6(a + b + c + 1)
The result is always divisible by 6, confirming the claim