Question 6 of 31intermediate🔧 ApplyShort Answer2 marks

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?

Correct Answer

Yes, Tathagat's claim is true.

Given: Numbers leaving remainder 2 when divided by 6 are of the form 6k+26k + 2.

Three such numbers: 6a+26a + 2, 6b+26b + 2, 6c+26c + 2

Sum:

(6a+2)+(6b+2)+(6c+2)(6a + 2) + (6b + 2) + (6c + 2) =6(a+b+c)+6= 6(a + b + c) + 6 =6(a+b+c+1)= 6(a + b + c + 1)

∴ The sum is always a multiple of 6.

Exercise: Figure it Out (Section 1) | Q: 6 | (Page Page 11)
For More Understanding

Explanation

The chapter discusses divisibility properties using algebra. Numbers leaving remainder 2 when divided by 6 have the general form 6k + 2. When three such numbers are added, the constant terms add up to 6, and the variable parts are all multiples of 6. The total sum is therefore always divisible by 6.

Solution Steps

  1. Numbers leaving remainder 2 when divided by 6: 6k + 2

  2. Three such numbers: 6a + 2, 6b + 2, 6c + 2

  3. Sum = 6a + 2 + 6b + 2 + 6c + 2

  4. Sum = 6(a + b + c) + 6

  5. Sum = 6(a + b + c + 1), always divisible by 6

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