Question 19 of 31intermediate🔍 AnalyzeLong Answer4 marks

When is the sum of two multiples of 3, a multiple of 6 and when is it not? Explain the different possible cases, and generalise the pattern.

Correct Answer

Given: Two multiples of 3: 3a and 3b

Sum: 3a + 3b = 3(a + b)

Condition for divisibility by 6: For 3(a + b) to be divisible by 6, we need a + b to be even.

Case 1: Both a and b are even Then 3a and 3b are multiples of 6. ∴ Sum is a multiple of 6.

Case 2: Both a and b are odd Then 3a and 3b are odd multiples of 3 (like 3, 9, 15). ∴ Sum is a multiple of 6.

Case 3: One even, one odd Then a + b is odd, so 3(a + b) is not divisible by 6. ∴ Sum is not a multiple of 6.

Generalisation: The sum of two multiples of 3 is a multiple of 6 if and only if both multiples are either even multiples of 3 (multiples of 6) or both are odd multiples of 3.

Exercise: Figure it Out (Section 4) | Q: 3 | (Page Page 21)
For More Understanding

Explanation

The chapter uses algebraic reasoning to analyze divisibility. Multiples of 3 can be classified as even multiples (6, 12, 18, ...) which are also multiples of 6, and odd multiples (3, 9, 15, ...) which are not multiples of 6. The sum of two even multiples of 3 is a multiple of 6. The sum of two odd multiples of 3 is also a multiple of 6 (since odd + odd = even). But the sum of one even and one odd multiple of 3 is not a multiple of 6.

Solution Steps

  1. Two multiples of 3: 3a and 3b

  2. Sum = 3(a + b)

  3. For divisibility by 6: need a + b to be even

  4. Case 1: Both a, b even → both multiples of 6 → sum multiple of 6

  5. Case 2: Both a, b odd → both odd multiples of 3 → sum multiple of 6

  6. Case 3: One even, one odd → a + b odd → sum not multiple of 6

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