Maths Worksheets Year 6 Problem Solving - Free Printable

Explanation:
We are given four numbers: 5, 6, 7, and 8.
We need to place them into four circles in a line (like A–B–C–D), so that the difference between each pair of joined numbers is more than one.

That means:
- |A − B| > 1
- |B − C| > 1
- |C − D| > 1

Let’s list all the possible differences between any two of these numbers:

- |5 − 6| = 1 → ✘ not allowed
- |5 − 7| = 2 → ✔ allowed
- |5 − 8| = 3 → ✔ allowed
- |6 − 7| = 1 → ✘ not allowed
- |6 − 8| = 2 → ✔ allowed
- |7 − 8| = 1 → ✘ not allowed

So the only *allowed adjacent pairs* are:
- (5,7), (7,5)
- (5,8), (8,5)
- (6,8), (8,6)

Note: (6,7) and (7,8) and (5,6) are NOT allowed — their difference is exactly 1.

We need a sequence of 4 distinct numbers using all of {5,6,7,8}, where each neighboring pair is from the allowed list above.

Let’s try building such a sequence.

Start with 5:
- From 5, allowed next numbers: 7 or 8.

Try 5 – 7:
- Now used: 5,7. Remaining: 6,8.
- From 7, allowed next numbers? Only numbers with difference >1 from 7:
- |7−5| = 2 ✔ but 5 already used
- |7−6| = 1 ✘
- |7−8| = 1 ✘
→ No valid next number after 7 (except 5, which is used). So 5–7 fails.

Try 5 – 8:
- Used: 5,8. Remaining: 6,7.
- From 8, allowed next numbers: must differ by >1 →
- |8−5| = 3 ✔ (used)
- |8−6| = 2 ✔ → OK
- |8−7| = 1 ✘
So next can be 6: 5 – 8 – 6
Now used: 5,8,6. Remaining: 7.
From 6, what’s left? Only 7.
|6 − 7| = 1 → ✘ Not allowed.

So 5–8–6 fails at the end.

What if after 5–8 we try something else? But only 6 and 7 remain, and only 6 is allowed after 8. So no.

Now try starting with 6:
- From 6, allowed next: only 8 (since |6−5|=1 ✘, |6−7|=1 ✘, |6−8|=2 ✔)

So 6 – 8
Used: 6,8. Remaining: 5,7
From 8, allowed next: 5 (|8−5|=3 ✔) or 6 (used). So try 6–8–5
Now used: 6,8,5. Remaining: 7
From 5, allowed next: 7 (|5−7|=2 ✔) → YES!

So sequence: 6 – 8 – 5 – 7

Check differences:
- |6 − 8| = 2 > 1 ✔
- |8 − 5| = 3 > 1 ✔
- |5 − 7| = 2 > 1 ✔

All good! All numbers used once.

Now the problem asks: “Can you put the numbers…?” → Yes, one way is 6, 8, 5, 7.

Then: “Now can you do it another way?”
Let’s find a second arrangement.

Try reverse of above: 7 – 5 – 8 – 6
Check:
- |7−5| = 2 ✔
- |5−8| = 3 ✔
- |8−6| = 2 ✔
Yes! That works too.

Another possibility: Start with 7?
From 7, allowed next: only 5 (|7−5|=2; others are 1). So 7–5
Then from 5: allowed next: 8 (|5−8|=3) → 7–5–8
Then remaining: 6 → |8−6|=2 ✔ → 7–5–8–6 — same as reversed earlier.

What about 8–6–5–7?
|8−6|=2 ✔
|6−5|=1 ✘ → invalid.

How about 8–5–7–6?
|8−5|=3 ✔
|5−7|=2 ✔
|7−6|=1 ✘ → invalid.

So only two valid sequences (and their reverses are the same two):
1. 6 – 8 – 5 – 7
2. 7 – 5 – 8 – 6

These are reverses of each other, but since the circles are in a line (order matters), both count as different arrangements.

So yes — one way is 6, 8, 5, 7. Another way is 7, 5, 8, 6.

The question just asks to provide arrangements — likely expects one for the first part, and another for the second.

Final Answer:
First arrangement: 6, 8, 5, 7
Second arrangement: 7, 5, 8, 6