## Poll

8 votes (47.05%) | |||

6 votes (35.29%) | |||

3 votes (17.64%) | |||

2 votes (11.76%) | |||

6 votes (35.29%) | |||

2 votes (11.76%) | |||

3 votes (17.64%) | |||

2 votes (11.76%) | |||

8 votes (47.05%) | |||

6 votes (35.29%) |

**17 members have voted**

July 23rd, 2020 at 8:00:02 AM
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Puzzle by Kevin J. Lin

It's Halloween and you've angered the spirit of Autumn by failing to revere the Great Pumpkin. Now a curse has befallen you. On the walkway to your house is a Ward of Seven Jack O' Lanterns arranged in a circle. If midnight comes and any of the seven are still lit, a dark reaper and seven dark horses with seven dark riders shall visit thy abode. They shall surround thy domicile and circle it seven times seven times. And the seven riders having circled thy dwelling seven times seven times, they shall proceed with the throwing of the eggs and the cream of shaving. And come morn there will be a great mess to be reckoned with. Verily.

So you'd better get those lanterns out.

You quickly discover something odd about these lanterns. When you blow out the first one, the lanterns on either side extinguish as well! But there's more- if you blow out a lantern adjacent to one that is extinguished, the extinguished one(s) will relight. It seems that blowing on any lantern will change the state of three- the one you blew on, and its two neighbors. Finally, you can blow on an extinguished lantern and it will relight, and its two neighbors will extinguish/ignite.

After a frustrating exercise of playing "Whack-A-Mole" with the lanterns, you find yourself back where you started, with all seven lit. Being the excellent puzzler you are, you sit down and examine the puzzle logically. But be quick, hoofsteps approach...

It's Halloween and you've angered the spirit of Autumn by failing to revere the Great Pumpkin. Now a curse has befallen you. On the walkway to your house is a Ward of Seven Jack O' Lanterns arranged in a circle. If midnight comes and any of the seven are still lit, a dark reaper and seven dark horses with seven dark riders shall visit thy abode. They shall surround thy domicile and circle it seven times seven times. And the seven riders having circled thy dwelling seven times seven times, they shall proceed with the throwing of the eggs and the cream of shaving. And come morn there will be a great mess to be reckoned with. Verily.

So you'd better get those lanterns out.

You quickly discover something odd about these lanterns. When you blow out the first one, the lanterns on either side extinguish as well! But there's more- if you blow out a lantern adjacent to one that is extinguished, the extinguished one(s) will relight. It seems that blowing on any lantern will change the state of three- the one you blew on, and its two neighbors. Finally, you can blow on an extinguished lantern and it will relight, and its two neighbors will extinguish/ignite.

After a frustrating exercise of playing "Whack-A-Mole" with the lanterns, you find yourself back where you started, with all seven lit. Being the excellent puzzler you are, you sit down and examine the puzzle logically. But be quick, hoofsteps approach...

Have you tried 22 tonight? I said 22.

July 23rd, 2020 at 10:07:10 AM
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Until I reread it, I didn't realise it was a ring of Seven lanterns.Luckily I had brought my dog (SCWT) along, so she saved me from leaning over as she found it easy to do all the blowing out for me!

It transpires by justing going 1,2,3,4,5,6,7 it gets rid of them all - you can also do it in any order. The logic is that every lantern has been affected three times, so has gone OFF,ON,OFF. By the time you finish every lantern is off.

July 23rd, 2020 at 4:16:59 PM
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Quote:charliepatrickIt transpires by justing going 1,2,3,4,5,6,7 it gets rid of them all - you can also do it in any order. The logic is that every lantern has been affected three times, so has gone OFF,ON,OFF. By the time you finish every lantern is off.

Correct!

So the Great Pumpkin has really given you a treat, not a trick, and the next morning all seven lanterns are full of candy.

Verily.

Verily.

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Q: What do you get if you divide the circumference of a pumpkin by its diameter?

A: Pumpkin pi.

Last edited by: Gialmere on Jul 23, 2020

Have you tried 22 tonight? I said 22.

July 26th, 2020 at 5:14:20 PM
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You've gotten lucky and have hit the bonus round on a dice themed slot machine.

The machine rolls a fair six-sided die on screen. You can stop and earn $100 dollars times the value of the roll, or you can roll again.

If you roll again, the same rules apply: you can stop and earn $100 dollars times the value of the roll, or you can roll again.

If you choose to roll a third time, you will earn $100 times the value of the roll and the round ends.

Assuming you use perfect strategy, what is the expected value of this bonus round?

Have you tried 22 tonight? I said 22.

July 26th, 2020 at 5:41:16 PM
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Gotta work backwards. The Expected value of the third roll is simply 100 x the average roll of a die, or 100 x 3.5 = 350.

Therefore, after the second roll, you stop if you roll a four, five or six, and re-roll if you roll a one to three.

So Expected value of the game at second roll = 1/6* 400 + 1/6 x 500 + 1/6 x 600 + 1/2 x 350 = 425.

Therefore on the first roll, stop with a 5 or 6, and roll with a 1 to 4.

So expected value on the first roll = expected value of the game = 1/6*500 + 1/6* 600 + 2/3 * 425 = 466.6666667

Therefore, after the second roll, you stop if you roll a four, five or six, and re-roll if you roll a one to three.

So Expected value of the game at second roll = 1/6* 400 + 1/6 x 500 + 1/6 x 600 + 1/2 x 350 = 425.

Therefore on the first roll, stop with a 5 or 6, and roll with a 1 to 4.

So expected value on the first roll = expected value of the game = 1/6*500 + 1/6* 600 + 2/3 * 425 = 466.6666667

July 26th, 2020 at 6:03:03 PM
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Quote:rsactuaryGotta work backwards. The Expected value of the third roll is simply 100 x the average roll of a die, or 100 x 3.5 = 350.

Therefore, after the second roll, you stop if you roll a four, five or six, and re-roll if you roll a one to three.

So Expected value of the game at second roll = 1/6* 400 + 1/6 x 500 + 1/6 x 600 + 1/2 x 350 = 425.

Therefore on the first roll, stop with a 5 or 6, and roll with a 1 to 4.

So expected value on the first roll = expected value of the game = 1/6*500 + 1/6* 600 + 2/3 * 425 = 466.6666667

Correct!

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"The diet book is one of those fool-and-money separation devices that seems, like roulette and slot machines, never to lose its power."

--Christopher Hitchens

Last edited by: Gialmere on Jul 26, 2020

Have you tried 22 tonight? I said 22.

July 27th, 2020 at 10:30:55 AM
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Five marbles of various sizes are placed in a conical funnel. Each marble is in contact with the adjacent marble(s). Also, each marble is in contact all around the funnel wall.

The smallest marble has a radius of 8mm. The largest marble has a radius of 18mm. What is the radius of the middle marble?

Have you tried 22 tonight? I said 22.

July 27th, 2020 at 11:33:53 AM
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12

I assume the ratio of the radiuses (what's the plural of radius and how do you spell it?) of two consecutive marbles is always the same. Let's call the radiuses, from bottom to top, 8, a, b, c, 18.

We have then:

a/8 = b/a = c/b = 18/c.

This leads to:

a^2 = 8b

b^2 = ac

c^2 = 18b

b^2 = sqrt(8b)*c

b^4 = 8b*c^2

b^3 = 8*c^2

b^3 = 8 * 18b

b^2 = 144

b = 12

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

July 27th, 2020 at 12:09:55 PM
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I had also guessed 12 assuming the ratios were the same. The ratio (18:middle) = ratio (middle:8). So that ratio^2 = 18/8 = 9/4. So that ratio = SQRT(9/4) = 3/2. So middle=12.

I can now see one way to prove it is by similar rectangles, adjacent to each other, creating four in a row such that the LHS of the first is 18 and the RHS of the fourth is 8.

Consider one of the funnel sides is on the x-axis and a circle, radius 18, lies resting on the x-axis at the origin (its centre at (0,18)). To the right, for arguments sake, is another circle which is slightly smaller; the second circle also lies resting on the x-axis and touches the first circle. Consider the rectangle with corners at the centres of each circle and the two points on the x-axis where the circles touch it. The rectangle has part of the x-asis, two edges at right angles and a slanted line through the centres of the two circles.

By creating a second similar rectangle, where the size of the larger circle is the same as the size of the smaller circle in the previous rectangle, the process can be continued until a row of five circles is created.

If the size of the second circle is created such that the first circle has radius 18 and the fifth circle radius 8, then this creates the "marbles" and shows the ratio between adjacent circles is constant.

I can now see one way to prove it is by similar rectangles, adjacent to each other, creating four in a row such that the LHS of the first is 18 and the RHS of the fourth is 8.

Consider one of the funnel sides is on the x-axis and a circle, radius 18, lies resting on the x-axis at the origin (its centre at (0,18)). To the right, for arguments sake, is another circle which is slightly smaller; the second circle also lies resting on the x-axis and touches the first circle. Consider the rectangle with corners at the centres of each circle and the two points on the x-axis where the circles touch it. The rectangle has part of the x-asis, two edges at right angles and a slanted line through the centres of the two circles.

By creating a second similar rectangle, where the size of the larger circle is the same as the size of the smaller circle in the previous rectangle, the process can be continued until a row of five circles is created.

If the size of the second circle is created such that the first circle has radius 18 and the fifth circle radius 8, then this creates the "marbles" and shows the ratio between adjacent circles is constant.

July 27th, 2020 at 4:19:29 PM
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Quote:Wizard12

I assume the ratio of the radiuses (what's the plural of radius and how do you spell it?) of two consecutive marbles is always the same. Let's call the radiuses, from bottom to top, 8, a, b, c, 18.

We have then:

a/8 = b/a = c/b = 18/c.

This leads to:

a^2 = 8b

b^2 = ac

c^2 = 18b

b^2 = sqrt(8b)*c

b^4 = 8b*c^2

b^3 = 8*c^2

b^3 = 8 * 18b

b^2 = 144

b = 12

Quote:charliepatrickI had also guessed 12 assuming the ratios were the same. The ratio (18:middle) = ratio (middle:8). So that ratio^2 = 18/8 = 9/4. So that ratio = SQRT(9/4) = 3/2. So middle=12.

I can now see one way to prove it is by similar rectangles, adjacent to each other, creating four in a row such that the LHS of the first is 18 and the RHS of the fourth is 8.

Consider one of the funnel sides is on the x-axis and a circle, radius 18, lies resting on the x-axis at the origin (its centre at (0,18)). To the right, for arguments sake, is another circle which is slightly smaller; the second circle also lies resting on the x-axis and touches the first circle. Consider the rectangle with corners at the centres of each circle and the two points on the x-axis where the circles touch it. The rectangle has part of the x-asis, two edges at right angles and a slanted line through the centres of the two circles.

By creating a second similar rectangle, where the size of the larger circle is the same as the size of the smaller circle in the previous rectangle, the process can be continued until a row of five circles is created.

If the size of the second circle is created such that the first circle has radius 18 and the fifth circle radius 8, then this creates the "marbles" and shows the ratio between adjacent circles is constant.

That is Mensa IQ Correct!

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In Maryland last weekend, two teenagers won the state marbles championship. The winning teens quickly said thank you, then boarded their time machine and returned to the year 1937.

Have you tried 22 tonight? I said 22.