Determine whether the statements below are 'Always True', 'Sometimes True', or 'Never True'. Explain your reasoning.
The product of a multiple of 6 and a multiple of 3 is a multiple of 9.
Always True. A multiple of 6 can be written as 6a and a multiple of 3 as 3b. Their product is 6a × 3b = 18ab = ab, which is always a multiple of 9.
The sum of three consecutive even numbers will be divisible by 6.
Always True. Three consecutive even numbers can be written as 2n, 2n+2, 2n+4. Their sum is 6n + 6 = 6(n + 1), which is always divisible by 6.
If abcdef is a multiple of 6, then badcef will be a multiple of 6.
Always True. A number is divisible by 6 if it is divisible by both 2 and 3. Divisibility by 3 depends on the sum of digits, which remains unchanged when digits are rearranged. Divisibility by 2 depends on the last digit being even. In both abcdef and badcef, the last digit is f (which is even since abcdef is divisible by 6). Therefore, badcef is also divisible by 6.
8(7b - 3) - 4(11b + 1) is a multiple of 12.
Never True. Expanding: 8(7b - 3) - 4(11b + 1) = 56b - 24 - 44b - 4 = 12b - 28 = 12(b - 2) - 4. This equals 12(b - 3) + 8, which is 8 more than a multiple of 12, never a multiple of 12 itself.
Explanation
(i) The chapter shows that the product of a multiple of 6 and a multiple of 3 always contains the factor 18 = , making it divisible by 9. (ii) The chapter demonstrates algebraic proof. Three consecutive even numbers always sum to a multiple of 6 because they form an arithmetic progression with common difference 2, and their sum is 3 times the middle term. (iii) The chapter explains that divisibility by 3 depends on digit sum (unchanged by rearrangement) and divisibility by 2 depends on the last digit (f in both cases). (iv) The chapter uses algebraic simplification. The expression simplifies to 12b - 28, which can be written as 12(b - 2) - 4 or 12(b - 3) + 8. In either form, it's clear the expression is never divisible by 12.