MathSphere Free Sample Maths Worksheets - Free Printable

Let's solve this step-by-step.

---

Problem Statement:

We are given the numbers: 5, 6, 7, and 8.
We need to place them in four circles connected in a line (like: ○ — ○ — ○ — ○), so that:

> The difference between each pair of joined numbers is more than one.

That means:
- For any two adjacent numbers, |A - B| > 1

We also need to find at least two different arrangements that satisfy this condition.

---

Step 1: Understand the constraint

We need to ensure that no two adjacent numbers differ by 1.

So let’s look at the differences between the numbers:

- |5 - 6| = 1 → Not allowed (difference is exactly 1)
- |6 - 7| = 1 → Not allowed
- |7 - 8| = 1 → Not allowed

So, adjacent numbers cannot be consecutive integers.

Therefore, we must avoid placing 5 next to 6, 6 next to 7, or 7 next to 8.

---

Step 2: Try to find valid arrangements

We have four positions:
Let’s label them: A — B — C — D

We need to assign 5, 6, 7, 8 to A, B, C, D such that:
- |A - B| > 1
- |B - C| > 1
- |C - D| > 1

Let’s try to build valid sequences.

---

#### Try 1: Start with 5

Try placing 5 in position A.

Then B cannot be 6 (since |5 - 6| = 1). So B can be 7 or 8.

Let’s try B = 7

Now check: |5 - 7| = 2 → OK

Now C cannot be 6 or 8? Let’s see:

- C must not be adjacent to B=7 → so C ≠ 6 and C ≠ 8 (because |7 - 6|=1, |7 - 8|=1)

But only numbers left are 6 and 8 → both are invalid!

So this path fails.

Try B = 8 instead.

|5 - 8| = 3 → OK

Now numbers left: 6 and 7

C must not be adjacent to 8 → so C ≠ 7 (since |8 - 7| = 1) → so C cannot be 7

So C must be 6 → then D = 7

Check:

- A = 5, B = 8 → |5 - 8| = 3 → OK
- B = 8, C = 6 → |8 - 6| = 2 → OK
- C = 6, D = 7 → |6 - 7| = 1 → ✘ Not allowed

So this fails.

So starting with 5 and putting 8 next doesn’t work.

Wait — maybe try 5 in the middle?

Let’s think differently.

---

Try a working arrangement

Let’s try: 5 — 7 — 6 — 8

Check differences:

- |5 - 7| = 2 → OK
- |7 - 6| = 1 → ✘ Not allowed

No good.

Try: 5 — 8 — 6 — 7

- |5 - 8| = 3 → OK
- |8 - 6| = 2 → OK
- |6 - 7| = 1 → ✘ No

Try: 6 — 8 — 5 — 7

- |6 - 8| = 2 → OK
- |8 - 5| = 3 → OK
- |5 - 7| = 2 → OK

All differences > 1 → ✔

So this works!

Let’s write it:

6 — 8 — 5 — 7

Check:
- 6 & 8 → diff = 2 → OK
- 8 & 5 → diff = 3 → OK
- 5 & 7 → diff = 2 → OK

Yes! Valid.

Now, let’s see if we can reverse it?

Try: 7 — 5 — 8 — 6

- |7 - 5| = 2 → OK
- |5 - 8| = 3 → OK
- |8 - 6| = 2 → OK

Also valid!

So another solution.

Wait — are these just reverses?

Let’s list both:

1. 6 — 8 — 5 — 7
2. 7 — 5 — 8 — 6

Are there more?

Let’s try: 5 — 8 — 7 — 6

- |5 - 8| = 3 → OK
- |8 - 7| = 1 → ✘ No

Try: 8 — 5 — 7 — 6

- |8 - 5| = 3 → OK
- |5 - 7| = 2 → OK
- |7 - 6| = 1 → ✘ No

Try: 5 — 7 — 8 — 6

- |5 - 7| = 2 → OK
- |7 - 8| = 1 → ✘ No

Try: 8 — 6 — 5 — 7

- |8 - 6| = 2 → OK
- |6 - 5| = 1 → ✘ No

Try: 7 — 8 — 5 — 6

- |7 - 8| = 1 → ✘ No

Try: 6 — 5 — 8 — 7

- |6 - 5| = 1 → ✘ No

Try: 8 — 7 — 6 — 5

- |8 - 7| = 1 → ✘ No

So far, only two seem to work.

Wait — what about: 5 — 8 — 6 — 7? Already tried → last pair diff = 1 → no

What about: 6 — 5 — 8 — 7? |6 - 5| = 1 → no

Try: 7 — 6 — 8 — 5

- |7 - 6| = 1 → ✘ No

Try: 8 — 6 — 7 — 5

- |8 - 6| = 2 → OK
- |6 - 7| = 1 → ✘ No

So only two seem to work.

Let’s list the two valid ones we found:

✔ 6 — 8 — 5 — 7
✔ 7 — 5 — 8 — 6

Are these different? Yes — they’re mirror images.

Let’s verify both:

First: 6 — 8 — 5 — 7
- 6 & 8 → 2 → OK
- 8 & 5 → 3 → OK
- 5 & 7 → 2 → OK

Second: 7 — 5 — 8 — 6
- 7 & 5 → 2 → OK
- 5 & 8 → 3 → OK
- 8 & 6 → 2 → OK

Perfect.

Is there a third?

Try: 5 — 7 — 6 — 8 → |5-7|=2, |7-6|=1 → ✘

Try: 8 — 5 — 6 — 7 → |8-5|=3, |5-6|=1 → ✘

Try: 6 — 7 — 8 — 5 → |6-7|=1 → ✘

Try: 5 — 6 — 7 — 8 → all adjacent differences = 1 → ✘

Try: 8 — 7 — 5 — 6 → |8-7|=1 → ✘

Try: 5 — 8 — 7 — 6 → |8-7|=1 → ✘

So only two valid arrangements.

Wait — what about 8 — 5 — 7 — 6?

- |8 - 5| = 3 → OK
- |5 - 7| = 2 → OK
- |7 - 6| = 1 → ✘ No

No.

How about 6 — 8 — 7 — 5?

- |6 - 8| = 2 → OK
- |8 - 7| = 1 → ✘ No

No.

So only two work.

---

✔ Final Answer:

Yes, we can do it. Here are two ways:

1. 6 — 8 — 5 — 7
Differences: |6-8|=2, |8-5|=3, |5-7|=2 → all >1 ✔

2. 7 — 5 — 8 — 6
Differences: |7-5|=2, |5-8|=3, |8-6|=2 → all >1 ✔

These are the only two possible arrangements (up to reversal).

---

Answer to the question:

Yes, you can put the numbers 5, 6, 7, and 8 into the circles so that the difference between each pair of joined numbers is more than one.

Here are two ways:

First way:
○ — ○ — ○ — ○
6 — 8 — 5 — 7

Second way:
○ — ○ — ○ — ○
7 — 5 — 8 — 6

Both satisfy the condition.