Question 22 of 31advanced🔧 ApplyShort Answer3 marks

If 3p7q8 is divisible by 44, list all possible pairs of values for p and q.

Correct Answer

Given: 3p7q8 is divisible by 44.

Rule: 44 = 4×114 \times 11, so the number must be divisible by both 4 and 11.

Divisibility by 4: The last two digits q8 must be divisible by 4. q8 ∈ {08, 28, 48, 68, 88} ∴ q ∈ {0, 2, 4, 6, 8}

Divisibility by 11: Alternating sum = 3 - p + 7 - q + 8 = 18 - p - q This must be divisible by 11. Since p and q are digits: 18 - p - q can equal 0 or 11.

Case 1: 18 - p - q = 0 p + q = 18 No valid solutions (with q even and p ≤\leq 9)

Case 2: 18 - p - q = 11 p + q = 7 With q ∈ {0, 2, 4, 6, 8}:

  • q = 0 → p = 7
  • q = 2 → p = 5
  • q = 4 → p = 3
  • q = 6 → p = 1
  • q = 8 → p = -1 (invalid)

Answer: The possible pairs (p, q) are: (7, 0), (5, 2), (3, 4), (1, 6)

Exercise: Figure it Out (Section 4) | Q: 6 | (Page Page 21)
For More Understanding

Explanation

The chapter discusses divisibility rules. For divisibility by 44 = 4×114 \times 11, the number must be divisible by both 4 and 11. For divisibility by 4, the last two digits q8 must be divisible by 4. This means q can be 0, 2, 4, 6, 8 (giving 08, 28, 48, 68, 88). For divisibility by 11, the alternating sum 3 - p + 7 - q + 8 = 18 - p - q must be divisible by 11. Since p and q are digits, 18 - p - q can equal 0 or 11. Case 1: 18 - p - q = 0 gives p + q = 18, but with q even and p ≤\leq 9, no valid solutions. Case 2: 18 - p - q = 11 gives p + q = 7. With q ∈ {0, 2, 4, 6, 8}: (p, q) = (7, 0), (5, 2), (3, 4), (1, 6).

Solution Steps

  1. 44 = 4×114 \times 11, need divisibility by both

  2. Divisible by 4: q8 divisible by 4 → q ∈ {0, 2, 4, 6, 8}

  3. Divisible by 11: 3 - p + 7 - q + 8 = 18 - p - q ≡ 0 (mod 11)

  4. 18 - p - q = 0 or 11

  5. Case 1: p + q = 18, no valid solutions with q even and p ≤\leq 9

  6. Case 2: p + q = 7, with q even → (p, q) = (7, 0), (5, 2), (3, 4), (1, 6)

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