Table mat
Feb 12, 7:05 PM. I've rephrased the question completely.
A table mat is made from red and white beads. There is an inner rectangle made entirely using white beads. It has a border (on all four sides) made from a single layer of red beads. Which mats can have as many white beads as red beads?
A table mat is made from red and white beads. There is an inner rectangle made entirely using white beads. It has a border (on all four sides) made from a single layer of red beads. Which mats can have as many white beads as red beads?
Labels: mathschallenge
posted by Chris at
2:29 PM
12 Comments:
Just to get in first, here's one solution: white beads 3 x 10 = 30, whole mat 5 x 12 = 60.
But I'm sure that there are many more solutions.
This post has been removed by the author.
This post has been removed by the author.
Successive multiples of 60 seem to yield many possibilities . . .
120 = 12 x 10 = 2 * (10 x 6)
180 = 15 x 12 = 2 * (10 x 9)
240 = 16 x 15 = 2 * (12 x 10)
300 = 20 x 15 = 2 * (15 x 10)
360 = 20 x 18 = 2 * (15 x 12)
. . . and so on and on.
The question doesn't state whether the width of the red beads has to be the same on opposite sides of the white beads rectangle, so some of these solutions give an asymmetrical pattern.
Overall mat would an area of (L*W).
Inner rectangular area of white beads (L₁*W₁). Area surrounding
(L₁*W₁) is red beads.
Desired mat should have these characteristics:
(L*W)-(L₁*W₁)= (L₁*W₁)
If I understand the description of the desire mat, that covers all the various possible dimensions.
Oooops! I should have said a single layer of red beads.
I'll amend the question.
In that case I'll add another solution to my first post:
White beads 6 x 4 = 24.
Red beads (6+2) x (4+2) = 48 = 2 x white beads.
Wiz has got two of the possible mats.
I think the only two.
Well done Wiz, you stood your ground like a champ. I'm sorry
that I, initially, messed up the question.
Let the inner white region be m by n and m and chooose m ≤ n.
The edge will have an area 2(m+1 + n+1) = 2m + 2n + 4.
Require mn = 2m + 2n + 4 => mn - 2m - 2n = 4 .
But (m - 2)(n - 2) = mn - 2m - 2n + 4.
So (m - 2)(n - 2) = 8. (m-2) and (n-2) must be factors of 8.
So (m-2) = 1 and (n-2) = 8 => m = 3, n = 10. So mat is 5*12
Or (m-2) = 2 and (n-2) = 4 => m = 4, n = 6. So mat is 6*8.
No other mat sizes will do it.
Thanks, Chris . . . this one deceptively difficult.
Still wrestling with Archimedes, Dedekind, et al . . .
As long as you're enjoying yourself. I really like that sort of stuff.
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