For each statement below, determine whether it is always true, sometimes true, or never true. Explain your answer. Mention examples and non-examples as appropriate. Justify your claim using algebra.
The sum of two even numbers is a multiple of 3.
Sometimes true. For example, 2 + 4 = 6 (multiple of 3), but 4 + 6 = 10 (not a multiple of 3). The sum of two even numbers is always even, but not necessarily a multiple of 3.
If a number is not divisible by 18, then it is also not divisible by 9.
Sometimes true. For example, 9 is divisible by 9 but not by 18. However, 10 is not divisible by either 9 or 18. A number can be divisible by 9 without being divisible by 18.
If two numbers are not divisible by 6, then their sum is not divisible by 6.
Sometimes true. For example, 4 and 2 are not divisible by 6, but 4 + 2 = 6 which is divisible by 6. However, 5 and 7 are not divisible by 6, and 5 + 7 = 12 which is divisible by 6. But 5 and 8 give 13 which is not divisible by 6.
The sum of a multiple of 6 and a multiple of 9 is a multiple of 3.
Always true. A multiple of 6 can be written as 6a and a multiple of 9 as 9b. Their sum is 6a + 9b = 3(2a + 3b), which is always a multiple of 3.
The sum of a multiple of 6 and a multiple of 3 is a multiple of 9.
Sometimes true. For example, 6 + 3 = 9 (multiple of 9), but 6 + 6 = 12 (not a multiple of 9). The sum depends on the specific values chosen.
Explanation
(i) The chapter discusses parity properties. Even numbers can be written as 2a and 2b. Their sum is 2(a+b), which is always even. Whether it's a multiple of 3 depends on whether a+b is a multiple of 3. (ii) The chapter explains that if A is divisible by k, then A is divisible by all factors of k. Since 9 is a factor of 18, numbers divisible by 18 are divisible by 9, but the reverse is not always true. (iii) The chapter discusses that if two numbers leave remainders that sum to 6 (or 0), their sum can be divisible by 6. Numbers like 4 and 2 leave remainders 4 and 2 respectively when divided by 6, and 4+2=6. (iv) The chapter demonstrates using algebra that if M and N are multiples of a number k, then M + N is also a multiple of k. Since both 6 and 9 are multiples of 3, their sum is also a multiple of 3. (v) A multiple of 6 is 6a and a multiple of 3 is 3b. Their sum is 6a + 3b = 3(2a + b). For this to be a multiple of 9, we need 2a + b to be a multiple of 3, which is not always the case.