Question 21 of 31intermediate🔧 ApplyShort Answer3 marks

If 48a23b is a multiple of 18, list all possible pairs of values for a and b.

Correct Answer

Given: 48a23b is a multiple of 18.

Rule: For divisibility by 18, the number must be divisible by both 2 and 9.

Divisibility by 2: b must be even: 0, 2, 4, 6, 8

Divisibility by 9: Sum of digits = 4 + 8 + a + 2 + 3 + b = 17 + a + b 17 + a + b must be divisible by 9. Since a and b are digits (0–9): 17 + a + b can equal 18 or 27.

Case 1: 17 + a + b = 18 a + b = 1 With b even: (a, b) = (1, 0)

Case 2: 17 + a + b = 27 a + b = 10 With b even: (a, b) = (2, 8), (4, 6), (6, 4), (8, 2)

∴ The possible pairs (a, b) are: (1, 0), (2, 8), (4, 6), (6, 4), (8, 2)

Exercise: Figure it Out (Section 4) | Q: 5 | (Page Page 21)
For More Understanding

Explanation

The chapter explains that for divisibility by 18, the number must be divisible by both 2 and 9. For divisibility by 2, b must be even (0, 2, 4, 6, 8). For divisibility by 9, the sum of digits 4 + 8 + a + 2 + 3 + b = 17 + a + b must be divisible by 9. Since a and b are digits (0-9), 17 + a + b can equal 18 or 27. Case 1: 17 + a + b = 18 gives a + b = 1. With b even: (a, b) = (1, 0). Case 2: 17 + a + b = 27 gives a + b = 10. With b even: (a, b) = (2, 8), (4, 6), (6, 4), (8, 2).

Solution Steps

  1. For divisibility by 18: must be divisible by 2 and 9

  2. Divisible by 2: b must be even (0, 2, 4, 6, 8)

  3. Divisible by 9: 17 + a + b must be divisible by 9

  4. 17 + a + b = 18 or 27 (since a, b ≤\leq 9)

  5. Case 1: a + b = 1, b even → (a, b) = (1, 0)

  6. Case 2: a + b = 10, b even → (a, b) = (2, 8), (4, 6), (6, 4), (8, 2)

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