Question 29 of 31intermediate🔍 AnalyzeLong Answer3 marks

Choose any 3 numbers. When is their sum divisible by 3? Explore all possible cases and generalise.

Correct Answer

Given: Three numbers, each leaving a remainder of 0, 1, or 2 when divided by 3.

Rule: The sum of three numbers is divisible by 3 when the sum of their remainders is divisible by 3 (i.e., 0, 3, or 6).

(1) Remainders (0, 0, 0): All three numbers are divisible by 3. Sum of remainders = 0 + 0 + 0 = 0 ∴ Sum is divisible by 3.

(2) Remainders (1, 1, 1): All three numbers leave the same non-zero remainder 1. Sum of remainders = 1 + 1 + 1 = 3 ∴ Sum is divisible by 3.

(3) Remainders (2, 2, 2): All three numbers leave the same non-zero remainder 2. Sum of remainders = 2 + 2 + 2 = 6 ∴ Sum is divisible by 3.

(4) Remainders (0, 1, 2): The three numbers have different remainders in any order. Sum of remainders = 0 + 1 + 2 = 3 ∴ Sum is divisible by 3.

Generalisation: Let the three numbers have remainders r₁, r₂, r₃ when divided by 3. The sum is divisible by 3 if and only if r₁ + r₂ + r₃ is divisible by 3.

Exercise: Figure it Out (Section 4) | Q: 13 | (Page Page 22)
For More Understanding

Explanation

The chapter explores divisibility using remainders. Any number divided by 3 leaves a remainder of 0, 1, or 2. For the sum to be divisible by 3, the sum of remainders must be 0, 3, or 6. This happens in the cases described above.

Solution Steps

  1. Each number divided by 3 leaves a remainder of 0, 1, or 2

  2. For the sum to be divisible by 3, the sum of remainders must be 0, 3, or 6

  3. Case (0, 0, 0): All three numbers are divisible by 3, sum of remainders = 0

  4. Case (1, 1, 1): All three numbers leave remainder 1, sum of remainders = 3

  5. Case (2, 2, 2): All three numbers leave remainder 2, sum of remainders = 6

  6. Case (0, 1, 2): Numbers have different remainders, sum of remainders = 3

  7. Generalisation: Sum is divisible by 3 if and only if r₁ + r₂ + r₃ is divisible by 3

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