## Poll

16 votes (50%) | |||

12 votes (37.5%) | |||

5 votes (15.62%) | |||

2 votes (6.25%) | |||

10 votes (31.25%) | |||

3 votes (9.37%) | |||

6 votes (18.75%) | |||

5 votes (15.62%) | |||

10 votes (31.25%) | |||

7 votes (21.87%) |

**32 members have voted**

November 27th, 2020 at 10:00:38 AM
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Today's puzzle is the simpler, 2d version of this week's toughie Tuesday question...

From the diagram above we can see that the number of tiles on the perimeter, 16, exceeds the number of tiles on the inside, 8.

How many rectangles exist for which the number of tiles on the perimeter are equal to the number of tiles on the inside?

From the diagram above we can see that the number of tiles on the perimeter, 16, exceeds the number of tiles on the inside, 8.

How many rectangles exist for which the number of tiles on the perimeter are equal to the number of tiles on the inside?

Have you tried 22 tonight? I said 22.

November 27th, 2020 at 11:13:54 AM
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Quote:GialmereHow many rectangles exist for which the number of tiles on the perimeter are equal to the number of tiles on the inside?

2

The following are the dimensions of the inner rectangle:

3x10

4x6

The following are the dimensions of the inner rectangle:

3x10

4x6

It's not whether you win or lose; it's whether or not you had a good bet.

November 27th, 2020 at 12:26:01 PM
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Let H and W be the dimensions, with H >= W

There are a total of HW squares, of which (H-2)(W-2) are interior

The number of interior = the number of exterior when HW = 2 (H-2)(W-2) = 2HW - 4H - 4W + 8

This is HW = 4H + 4W - 8

H (W - 4) = 4W - 8

H = (4W - 8) / (W - 4)

H = (4W - 16 + 8) / (W - 4)

H = 4 + 8 / (W - 4)

Since H is an integer, (W - 4) is a factor of 8

The only factors of 8 are 1, 2, 4, and 8, with resulting H values of 12, 8, 6, and 5, but the latter two have W > H

The only H values where H > W are 12 and 8; the corresponding W values are 5 and 6

12 x 5 and 8 x 6 are the only solutions

November 27th, 2020 at 8:45:18 PM
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Quote:Wizard2

The following are the dimensions of the inner rectangle:

3x10

4x6

Quote:ThatDonGuy

Let H and W be the dimensions, with H >= W

There are a total of HW squares, of which (H-2)(W-2) are interior

The number of interior = the number of exterior when HW = 2 (H-2)(W-2) = 2HW - 4H - 4W + 8

This is HW = 4H + 4W - 8

H (W - 4) = 4W - 8

H = (4W - 8) / (W - 4)

H = (4W - 16 + 8) / (W - 4)

H = 4 + 8 / (W - 4)

Since H is an integer, (W - 4) is a factor of 8

The only factors of 8 are 1, 2, 4, and 8, with resulting H values of 12, 8, 6, and 5, but the latter two have W > H

The only H values where H > W are 12 and 8; the corresponding W values are 5 and 6

12 x 5 and 8 x 6 are the only solutions

Correct!

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The Wi-Fi at my house went out this Thanksgiving.

There are 148 tiles in my bathroom.

Have you tried 22 tonight? I said 22.

November 30th, 2020 at 8:35:38 AM
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I have some difficult puzzles this week so even easy Monday isn't all that easy...

You go to visit your pool playing friend and shoot a few games. He's a very superstitious player and had a special 16-ball made to replace his table's 8-ball.

He also insists on racking the balls in a certain way. He arranges them so that each ball is the difference of the two balls above it. He begins to show you how to do it when his phone rings and he leaves the room.

Can you figure out how to finish racking the balls?

You go to visit your pool playing friend and shoot a few games. He's a very superstitious player and had a special 16-ball made to replace his table's 8-ball.

He also insists on racking the balls in a certain way. He arranges them so that each ball is the difference of the two balls above it. He begins to show you how to do it when his phone rings and he leaves the room.

Can you figure out how to finish racking the balls?

Have you tried 22 tonight? I said 22.

November 30th, 2020 at 10:29:05 AM
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Quote:GialmereCan you figure out how to finish racking the balls?

It's not so much hard as exhaustive, as it's a bit of a brute force problem, but the solution is:

5, 14, 16, 3, 15

9, 2, 13, 12

7, 11, 1

4, 10

6

November 30th, 2020 at 5:08:12 PM
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Quote:ThatDonGuy

It's not so much hard as exhaustive, as it's a bit of a brute force problem, but the solution is:

5, 14, 16, 3, 15

9, 2, 13, 12

7, 11, 1

4, 10

6

Correct!

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Have you tried 22 tonight? I said 22.

December 1st, 2020 at 9:04:07 AM
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It's toughie Tuesday...

A 5 by 5 square lattice is formed by drilling holes in a piece of wood. Three pegs are placed in this lattice at random.

What is the probability that the three pegs will form a triangle?

A 5 by 5 square lattice is formed by drilling holes in a piece of wood. Three pegs are placed in this lattice at random.

What is the probability that the three pegs will form a triangle?

Have you tried 22 tonight? I said 22.

December 1st, 2020 at 9:31:11 AM
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Total perms =25*24*23/6 = 2208

For a triangle to have zero area the holes must be in the same line, diagonal or horizontal/vertical.

Diagonal 3-length lines - 4 of them, 1 perm = 4 combos

Diagonal 4-length lines - 4 of them, 4 perms = 16 combos

Diagonal 5-length lines - 2 of them, 10 perms = 20 combos

Vertical/Horizontal 5-length lines - 10 of them, 10 perms = 100 combos

Total = 140 combos.

So P=140/2208.

For a triangle to have zero area the holes must be in the same line, diagonal or horizontal/vertical.

Diagonal 3-length lines - 4 of them, 1 perm = 4 combos

Diagonal 4-length lines - 4 of them, 4 perms = 16 combos

Diagonal 5-length lines - 2 of them, 10 perms = 20 combos

Vertical/Horizontal 5-length lines - 10 of them, 10 perms = 100 combos

Total = 140 combos.

So P=140/2208.

Last edited by: charliepatrick on Dec 1, 2020

December 1st, 2020 at 10:21:09 AM
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Nice work! Is anyone else having a difficult time making sense of this form?

Math is the only true form of knowledge