4d - 2o = 10a + b

4d - 3o = 10b + a ( a and b are arbitrary numbers between 1 and 9 inclusive)

Quite a lot of this cancels out, leaving o = 9(a-b).

(If o=9 then a-b=1, a>b; o<=d/2)

I came up with almost the same equations but yours are neater (using d instead of r) so I quote yours, hope you don't mind Mathsyperson.

I tried to cancel o out, leaving 4d=28a-17b

4d is even so 17b must be even or b even.

d, a, b are integers and d>0, a>0, b>0, a<>b

Replace a, b with value 1..9 or 2..8 and I have d=18; 8; 32; 46 mm

d=8 too small, d=46 too big a coin, that left d=18 mm (a=5, b=4) or 32 mm (a=7, b=4).

But it looks like only d=18 mm works.

What's wrong with my solution ?

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So which is your FINAL answer out of the two?

I will post my solution soon.

In the mean time, can you tell me what the overlap is given your coin diameters? That should give you a clue

]]>And if that's not right then I'm well and truly stumped.

]]>I can only see one way of doing this problem, so I would be interested in what other methods can be used to work this out.

I will submit my solution once there is more discussion!

]]>If I stacked the other three coins in the tray in a mirror image pattern of the first three, the smallest tray would be a two-figure number of millimetres wide. But if I repeated the above pattern, putting in a single coin next, then two on the right, the trays width could be reduced. In fact, it would be the same as before but with its two digits reversed.

**What is the diameter of each coin?**