Given the information:

1. The highest common factor (HCF) of two numbers $x$ and $y$ is 7.
2. Their sum is 56, so $x + y = 56$.
3. We are asked for the number of distinct values of $x - y$ that are possible.

Step-by-Step Solution

Since the HCF of $x$ and $y$ is 7, both $x$ and $y$ must be multiples of 7. Let: $x = 7a \quad \text{and} \quad y = 7b$ where $a$ and $b$ are integers that are coprime (since 7 is their HCF).

Substituting $x = 7a$ and $y = 7b$ into $x + y = 56$, we get: $7a + 7b = 56$ Dividing by 7: $a + b = 8$

We now need to find possible integer values of $a$ and $b$ such that $a + b = 8$ and $a > b$.

Possible Pairs $(a, b)$ Satisfying $a + b = 8$

The pairs $(a, b)$ with $a > b$ are:

1. $(7, 1)$
2. $(6, 2)$
3. $(5, 3)$

Calculating $x - y$ for Each Pair

1. For $(a, b) = (7, 1)$: $x - y = 7a - 7b = 7 \times 7 - 7 \times 1 = 49 - 7 = 42$

2. For $(a, b) = (6, 2)$: $x - y = 7a - 7b = 7 \times 6 - 7 \times 2 = 42 - 14 = 28$

3. For $(a, b) = (5, 3)$: $x - y = 7a - 7b = 7 \times 5 - 7 \times 3 = 35 - 21 = 14$

Distinct Values of $x - y$

The distinct values of $x - y$ are 42, 28, and 14. So, there are 3 distinct values of $x - y$ possible.

Answer

The correct answer is: $\boxed{3}$

Would you like further details or have any questions?

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1. Why must $x$ and $y$ be multiples of their HCF?
2. Can the values of $x$ and $y$ be non-integers in this type of problem?
3. How can we confirm $x > y$ based on the pairs $(a, b)$?
4. Why is it necessary to ensure $a$ and $b$ are coprime?
5. What are other methods to find distinct values of $x - y$?

Tip: When working with HCF problems, always check if the numbers can be written as multiples of the HCF to simplify the problem.