Easy math puzzles
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46 members have voted
July 22nd, 2020 at 8:02:25 AM
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Quote: charliepatrickI've have to resort to a spreadsheet but can see that the correct strategy is for Player #1 to look through 100/e (=37) and after that pick the box that's the higher than any seen so far. Probability of success = .371043 (slightly higher than 1/e).
So the second player has two options...
Scenario A - Player #2 looks at the first N/e = 37 then picks the second box higher than that seen so far (he knows the first one encountered wasn't highest).
Scenario B - Player #2 assumes the best box is in the first 37 and uses the same strategy for just those - Pr success ~= 1/e * 37/100 = .137286
(i) Player #1 - finds the best in the first 37 boxes, so cannot win (and therefore goes right through the boxes). Pr = 37/100
(This is the only case where the audience can help, let's assume they don't - so #2 will also lose.)
(ii) Player #1 - finds the 2nd best in the first N, only 1st is left, so will always win.
(iii) Player #1 - finds the 3rd best in the first N, so 1st and 2nd left, so has a 50/50 chance to win; if lose the 2nd competitor will always win.
(iv) Player #1 - finds the 4th best in the first N, 1st 2nd 3rd left, so has 1/3 chance to win, if lose the 2nd competitor has a 1/2 chance to win.
(lxx) 1-70 left, so has 1/70 chance to win, if lose the 2nd competitor has a 1/69 chance to win.
Assuming Player No2 adopts this strategy, the probabiity without knowing than Player No1 didn't win, is ....
Pr #1 or #2 wins if the 1st is in the first half = 0.
Pr #1 wins = Pr(finds the 2nd in the first 37)*1 + Pr(finds the 3rd in the first 37)*1/2 + Pr(finds the 4th in the first 37)*1/3 etc.
Pr #2 wins = Pr(finds the 2nd in the first 37)*0 + Pr(finds the 3rd in the first 37)*1 + Pr(finds the 4th in the first 37)*1/2 etc.
This works out to be .233740 (which is close to 1/e/(1-1/e)).
However the above includes situations where Player #1 would win. e.g. if the 2nd highest was in the first 37, Player No2 wouldn't be playing. So we have to exclude cases where Player #1 won. This is 1/e, so we have to multiply the probability by (1-1/e).
Gosh this means the chances that Player #2 wins as (.371631) about 1/e!!!
If the audience helps then you can add best box was in the first 37, and that's 1/e as well, so gives 2/e.
Isn't Life Strange - Moody Blues
Correct!
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My girlfriend left because I treated our relationship like a game show...
Oh well, she was a worthy contestant
Have you tried 22 tonight? I said 22.
July 22nd, 2020 at 8:05:03 AM
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Could somebody check my math here? I think I found how to determine the solution, but I'm not quite sure I analyzed it correctly.
Let N be the number of secretaries - in our case, 100
Let S be the number at which you skip the first (S-1) secretaries, then choose the first one with a salary/ranking higher than the earlier ones (obviously, you don't choose a secretary that would be lower than an earlier one, as your selection would be incorrect)
The probability that secretary S is correct and is chosen is:
SUM(k=1..N) {P(#k is correct and is chosen)}
= SUM(k=1..N) {P(#k is correct) P(#k is chosen given that #k is correct)}
= SUM(k=S..N) {1/N P(#k is chosen given that #k is correct)}
(note P(#k is chosen given that #k is correct) = 0 for k = 1..S-1)
= SUM(k=S..N) {1/N P("the best of the first (k-1) secretaries is in the first (S-1)" given that #k is correct)}
(because if the best of the first (k-1) secretaries is not in the first (S-1), then that secretary will be chosen before #k)
= SUM(k=S..N) {1/N x (S-1)/(k-1)}
= (S-1) / N x SUM(k=S..N) {1/(k-1)}
The answer is the value of S where this is a maximum.
From Thomas Ferguson's article in Statistical Science, the general solution S = N/e apparently comes from the fact that, if x = LIMIT(n -> +INF) {r/n}, then, setting t = k/n (and dt =1/n), the sum becomes a Riemann approximation to the integral
x INTEGRAL(x, 1) {1/t dt}, which is -x ln x; the first derivative of -x ln x = (-1 - ln x), which is 0 when ln x = -1 -> x = 1/e -> -x ln x = -1/e * -1 = 1/e.
Let N be the number of secretaries - in our case, 100
Let S be the number at which you skip the first (S-1) secretaries, then choose the first one with a salary/ranking higher than the earlier ones (obviously, you don't choose a secretary that would be lower than an earlier one, as your selection would be incorrect)
The probability that secretary S is correct and is chosen is:
SUM(k=1..N) {P(#k is correct and is chosen)}
= SUM(k=1..N) {P(#k is correct) P(#k is chosen given that #k is correct)}
= SUM(k=S..N) {1/N P(#k is chosen given that #k is correct)}
(note P(#k is chosen given that #k is correct) = 0 for k = 1..S-1)
= SUM(k=S..N) {1/N P("the best of the first (k-1) secretaries is in the first (S-1)" given that #k is correct)}
(because if the best of the first (k-1) secretaries is not in the first (S-1), then that secretary will be chosen before #k)
= SUM(k=S..N) {1/N x (S-1)/(k-1)}
= (S-1) / N x SUM(k=S..N) {1/(k-1)}
The answer is the value of S where this is a maximum.
From Thomas Ferguson's article in Statistical Science, the general solution S = N/e apparently comes from the fact that, if x = LIMIT(n -> +INF) {r/n}, then, setting t = k/n (and dt =1/n), the sum becomes a Riemann approximation to the integral
x INTEGRAL(x, 1) {1/t dt}, which is -x ln x; the first derivative of -x ln x = (-1 - ln x), which is 0 when ln x = -1 -> x = 1/e -> -x ln x = -1/e * -1 = 1/e.
July 22nd, 2020 at 11:35:40 AM
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Quote: ThatDonGuyCould somebody check my math here? I think I found how to determine the solution, but I'm not quite sure I analyzed it correctly.
Let N be the number of secretaries - in our case, 100
Let S be the number at which you skip the first (S-1) secretaries, then choose the first one with a salary/ranking higher than the earlier ones (obviously, you don't choose a secretary that would be lower than an earlier one, as your selection would be incorrect)
The probability that secretary S is correct and is chosen is:
SUM(k=1..N) {P(#k is correct and is chosen)}
= SUM(k=1..N) {P(#k is correct) P(#k is chosen given that #k is correct)}
= SUM(k=S..N) {1/N P(#k is chosen given that #k is correct)}
(note P(#k is chosen given that #k is correct) = 0 for k = 1..S-1)
= SUM(k=S..N) {1/N P("the best of the first (k-1) secretaries is in the first (S-1)" given that #k is correct)}
(because if the best of the first (k-1) secretaries is not in the first (S-1), then that secretary will be chosen before #k)
= SUM(k=S..N) {1/N x (S-1)/(k-1)}
= (S-1) / N x SUM(k=S..N) {1/(k-1)}
The answer is the value of S where this is a maximum.
From Thomas Ferguson's article in Statistical Science, the general solution S = N/e apparently comes from the fact that, if x = LIMIT(n -> +INF) {r/n}, then, setting t = k/n (and dt =1/n), the sum becomes a Riemann approximation to the integral
x INTEGRAL(x, 1) {1/t dt}, which is -x ln x; the first derivative of -x ln x = (-1 - ln x), which is 0 when ln x = -1 -> x = 1/e -> -x ln x = -1/e * -1 = 1/e.
As an aside, we can see from ThatDonGuy's response that digital social media will be changing mathematical notation, probably forever.
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
July 22nd, 2020 at 12:43:37 PM
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The towns of Alpha, Beta, and Gamma are equidistant from each other.
You're driving to visit a friend in Gamma but have gotten lost. You come across a signpost stating that you are three miles from Alpha and four miles from Beta.
What then is your maximum possible distance from Gamma? (Assume the land is flat.)
Have you tried 22 tonight? I said 22.
July 22nd, 2020 at 2:12:43 PM
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Quote: Gialmere
The towns of Alpha, Beta, and Gamma are equidistant from each other.
You're driving to visit a friend in Gamma but have gotten lost. You come across a signpost stating that you are three miles from Alpha and four miles from Beta.
What then is your maximum possible distance from Gamma? (Assume the land is flat.)
Surprisingly, 7 miles. If you are the 7 miles from Gamma, then the distance between cities is about 6.0825. I don't see the logic that makes this an "easy" puzzle, though.
I heart Crystal Math.
July 22nd, 2020 at 3:43:17 PM
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Using a spreadsheet I agree, but also haven't yet seen the logic. btw I get a very slightly different figure...Quote: CrystalMathSurprisingly, 7 miles. If you are the 7 miles from Gamma, then the distance between cities is about 6.0825. I don't see the logic that makes this an "easy" puzzle, though.
6.082 763 | X - axis | Y - axis | ||
A (aka Alpha) | 0.000 000 | 0.000 000 | ||
B (aka Beta) | 6.082 763 | 0.000 000 | ||
C (aka Gamma) | 3.041 381 | -5.267 827 | ||
Me | 2.465 985 | 1.708 484 | 0.904 194 | Cos C |
Dist - Deltas | -0.575 396 | 6.976 311 | 7.000 000 |
July 22nd, 2020 at 3:50:19 PM
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Quote: CrystalMath
Surprisingly, 7 miles. If you are the 7 miles from Gamma, then the distance between cities is about 6.0825. I don't see the logic that makes this an "easy" puzzle, though.
Quote: charliepatrick
Correct!
Quote: CrystalMathI don't see the logic that makes this an "easy" puzzle, though.
You have a point. This one just looks easy.
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I want to die peacefully in my sleep like my grandfather, but not like those screaming passengers in the car with him.
Have you tried 22 tonight? I said 22.
July 22nd, 2020 at 4:46:08 PM
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Good problem. I don't immediately see why the answer is what CM said, not that I'm disputing it.
Reminder that math puzzles don't have to be posted here -- just easy ones that are not thread worthy. This one is thread worthy.
Reminder that math puzzles don't have to be posted here -- just easy ones that are not thread worthy. This one is thread worthy.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
July 22nd, 2020 at 6:52:16 PM
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Just had a try with the sides 4 and 4 (rather than 3 and 4) and guess what the answer is!!
Yes you've guessed it, the answer is 8.
What's more interesting is the distance AB = SQRT(48) and, if AB is the x-axis, I am 2 above and G is 6 below. (Note that if the cities are ABG and I'm at M, AG^2+AM^2=SQRT(48)+4*4=8*8=GM^2; so the angle GAM is 90 degrees.)
Using the spreadsheet I can also see a pattern emerging (the length is always the sum of the two)...
0 4 SQRT(16)
1 4 SQRT(21) +5
2 4 SQRT(28) +7
3 4 SQRT(37) +9
4 4 SQRT(48) +11
4 5 SQRT(61)
3 6 SQRT(63) +2
2 7 SQRT(67) +4
1 8 SQRT(73) +6
0 9 SQRT(81) +8
The pattern can be found if you consider Cos(angle)=(a^2+b^2-c^2)/(2ab) as all the angles above are 120 degrees.
So all one has to do is show the angle is 120 degrees creates the maximum answer and this means length will be SQRT(a^2+b^2+ab).
If you carry on this logic you see the angles AMG and BMG both have to be 60 degrees - e.g. Cos(Theta)=(adj^2+adj^2-opp^2)/(2 adj adj) = 49+9-37)/(2*3*7) = 1/2 or (16+49-37)/(2*4*7) = 1/2.
What's more interesting is the distance AB = SQRT(48) and, if AB is the x-axis, I am 2 above and G is 6 below. (Note that if the cities are ABG and I'm at M, AG^2+AM^2=SQRT(48)+4*4=8*8=GM^2; so the angle GAM is 90 degrees.)
Using the spreadsheet I can also see a pattern emerging (the length is always the sum of the two)...
0 4 SQRT(16)
1 4 SQRT(21) +5
2 4 SQRT(28) +7
3 4 SQRT(37) +9
4 4 SQRT(48) +11
4 5 SQRT(61)
3 6 SQRT(63) +2
2 7 SQRT(67) +4
1 8 SQRT(73) +6
0 9 SQRT(81) +8
The pattern can be found if you consider Cos(angle)=(a^2+b^2-c^2)/(2ab) as all the angles above are 120 degrees.
So all one has to do is show the angle is 120 degrees creates the maximum answer and this means length will be SQRT(a^2+b^2+ab).
If you carry on this logic you see the angles AMG and BMG both have to be 60 degrees - e.g. Cos(Theta)=(adj^2+adj^2-opp^2)/(2 adj adj) = 49+9-37)/(2*3*7) = 1/2 or (16+49-37)/(2*4*7) = 1/2.
Last edited by: charliepatrick on Jul 22, 2020
July 22nd, 2020 at 8:24:09 PM
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btw I wonder whether this is connected to a puzzle where there are three cities and you want to join them by roads. The plan is to find a point in the middle for a junction and then build a road from each city to the junction.
Apart from silly cases, where they're in a straight line, or it's quicker to have the junction at one of the cities, you can prove the angles the roads meet at the junction are all 120 degrees. Some of the logic is based on symmetry that the forces must be equal. Thus I was wondering whether one can use this idea or to construct a point on GM extended.
Yes you can!
Consider locating a city D just Northish of "Me" on the line GM at a fixed distance from G. Then find the minimum "roads" joining A B and D. Since AM and BM are fixed, in this case 3 miles and 4 miles, we are minimizing DM. This therefore maximises MC.
Yes you can!
Consider locating a city D just Northish of "Me" on the line GM at a fixed distance from G. Then find the minimum "roads" joining A B and D. Since AM and BM are fixed, in this case 3 miles and 4 miles, we are minimizing DM. This therefore maximises MC.
July 23rd, 2020 at 8:00:02 AM
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Puzzle by Kevin J. Lin
/pic5546229.png)
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.
-------------------------------
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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/pic5547733.png)
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
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
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.
July 27th, 2020 at 5:48:01 PM
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Quote: Wizardwhat's the plural of radius and how do you spell it?
Radii (I pronounce it "RAID-ee-eye")
July 27th, 2020 at 6:29:51 PM
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What’s the plural of Lexus ?
It’s all about making that GTA
July 27th, 2020 at 8:02:49 PM
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Quote: ChesterDogHere are some rules about irregular plurals.
Try to figure out the singular of "latices." (Recently, I saw the word "latices" and had to look it up.)
...that "latex" is not a proper noun. Remind me not to challenge it if somebody uses it in Scrabble.
July 28th, 2020 at 4:45:53 AM
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I didn't know the plural of penny was pence.
Now that I think about it, I think I've heard "pence" on UK television, but thought it meant more like "cents."
Now that I think about it, I think I've heard "pence" on UK television, but thought it meant more like "cents."
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
July 28th, 2020 at 5:29:03 AM
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Prior to decimalisation (1971) we had Pounds, Shillings and Pence. So there are also expressions thruppence, a threepenny bit, etc..
The problem I have is with deer, fish, sheep etc. You can never tell how many there are!
The problem I have is with deer, fish, sheep etc. You can never tell how many there are!
July 28th, 2020 at 7:26:37 AM
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Quote: charliepatrickPrior to decimalisation (1971) we had Pounds, Shillings and Pence. So there are also expressions thruppence, a threepenny bit, etc..
What about a guinea?
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
July 28th, 2020 at 9:08:53 AM
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Puzzle by Shakuntala Devi
Fifty minutes ago if it was four times as many minutes past three o'clock, how many minutes is it to six o'clock?
Fifty minutes ago if it was four times as many minutes past three o'clock, how many minutes is it to six o'clock?
Have you tried 22 tonight? I said 22.
July 28th, 2020 at 10:48:01 AM
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650 Minutes.
50 minutes ago it was 200 past 3 or 6:20. So now, 50 minutes later, it is 7:10. In 10 hours & 50 minutes it will be 6:00. BTW, you can add 720 minutes to that total if you are using a 24-hour clock.
50 minutes ago it was 200 past 3 or 6:20. So now, 50 minutes later, it is 7:10. In 10 hours & 50 minutes it will be 6:00. BTW, you can add 720 minutes to that total if you are using a 24-hour clock.
"Dealer has 'rock'... Pay 'paper!'"
July 28th, 2020 at 11:07:33 AM
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Quote: Joeman650 Minutes.
50 minutes ago it was 200 past 3 or 6:20. So now, 50 minutes later, it is 7:10. In 10 hours & 50 minutes it will be 6:00. BTW, you can add 720 minutes to that total if you are using a 24-hour clock.
Sorry, no.
It's actually simpler than that but the crafty Devi confuses things with the way she phrases the question.
Have you tried 22 tonight? I said 22.
July 28th, 2020 at 11:20:04 AM
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Quote: Gialmere
Fifty minutes ago if it was four times as many minutes past three o'clock, how many minutes is it to six o'clock?
Very tricky wording. I would have first given the same answer as Joeman.
26 minutes to 6:00 or 5:34
50 minutes ago it was 4:44, which is 104 minutes past 3:00. 104 minutes is 4*26.
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
July 28th, 2020 at 11:43:03 AM
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A drunk takes a step south with probability p (where 0<p<0.5) and otherwise takes a step north. What is the probability he ever returns to the point he started?
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
July 28th, 2020 at 11:49:27 AM
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Quote: gordonm888Very tricky wording. I would have first given the same answer as Joeman.
26 minutes to 6:00 or 5:34
50 minutes ago it was 4:44, which is 104 minutes past 3:00. 104 minutes is 4*26.
Correct!
------------------------------------
A film about the life of Shakuntala Devi (aka The Human Computer) was scheduled for theater release this year but got covid canceled. It will instead start streaming on Amazon this Friday.
Have you tried 22 tonight? I said 22.
July 28th, 2020 at 12:13:36 PM
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Quote: WizardA drunk takes a step south with probability p (where 0<p<0.5) and otherwise takes a step north. What is the probability he ever returns to the point he started?
I am not entirely sure why, but I get 2p
The value is 2 times the sum of (p (1-p) + the sum over all positive integers n of ( (p (1-p))^(n+1) x (Combin(2n, n) - Combin(2n, n-1)) )
July 28th, 2020 at 12:25:22 PM
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The way the question is written, I still stand by my answer. To get her answer, the question would have been better written as: "How many minutes is it to six o'clock? Fifty minutes ago if it was four times as many minutes past three o'clock."Quote: GialmereQuote: gordonm888Very tricky wording. I would have first given the same answer as Joeman.
26 minutes to 6:00 or 5:34
50 minutes ago it was 4:44, which is 104 minutes past 3:00. 104 minutes is 4*26.
Correct!
Ambiguity in a contract benefits the party that did not draft it. I learned that watching The Big Bang Theory! ;)
"Dealer has 'rock'... Pay 'paper!'"
July 28th, 2020 at 1:02:10 PM
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I agree with JoeMan.Quote: JoemanThe way the question is written, I still stand by my answer. To get her answer, the question would have been better written as: "How many minutes is it to six o'clock? Fifty minutes ago if it was four times as many minutes past three o'clock."Quote: GialmereQuote: gordonm888Very tricky wording. I would have first given the same answer as Joeman.
26 minutes to 6:00 or 5:34
50 minutes ago it was 4:44, which is 104 minutes past 3:00. 104 minutes is 4*26.
Correct!
Ambiguity in a contract benefits the party that did not draft it. I learned that watching The Big Bang Theory! ;)
An English language judge might be needed
It’s all about making that GTA
July 28th, 2020 at 1:28:03 PM
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Quote: WizardA drunk takes a step south with probability p (where 0<p<0.5) and otherwise takes a step north. What is the probability he ever returns to the point he started?
p/(1-p)
It’s all about making that GTA
July 28th, 2020 at 4:15:31 PM
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Any more?
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
July 29th, 2020 at 9:26:06 AM
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/pic5555629.png)
The rectangle above is divided into 9 perfect squares.
The smallest square (colored black) has a side length of 4cm.
What is the area of the rectangle?
Have you tried 22 tonight? I said 22.
July 29th, 2020 at 10:07:01 AM
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Quote: GialmereThe rectangle above is divided into 9 perfect squares.
The smallest square (colored black) has a side length of 4cm.
What is the area of the rectangle?
Label the eight squares as follows:
A - upper left
B - upper right
C - left edge, below A
D - touching A, B, C
E - below D
F - bottom left
G - bottom center
H - bottom right
G = E + 4
F = G + 4 = E + 8
C = F + 4 = E + 12
C + 4 = D + E -> E + 16 = D + E -> D = 16
A = C + D = E + 12 + 16 = E + 28
B = A + D = E + 44
B + D = E + H -> E + 44 + 16 = E + H -> H = 60
H = E + G = 2E + 4 -> E = 28
The eight squares A, B, C, D, E, F, G, H have side lengths 56, 72, 40, 16, 28, 36, 32, and 60, respectively
The width of the rectangle = A + B = 128
Check: it is also F + G + H = 128
The height of the rectangle = A + C + F = 132
Check: it is also B + H = 132
The area = 128 x 132 = 16,896
July 29th, 2020 at 10:32:08 AM
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If you call the squares starting at the top, A B, C D, E F, G H. Then if you start at the square NE of the small square (E). Then work clockwise.
(i) H=E+4, consider its top edge which meet "4" and "E".
(ii) G=H+4=E+8, consider its Eastern edge which meet "4" and "H".
(iii) Similarly C=G+4=E+12.
(iv) The height of C and G is the same as D, E and H. (E+12)+(E+8)=D+E+(E+4). So D=16.
(v) A= C+D = (E+12)+16 = E+28.
(vi) B= A+C = (E+28)+16 = E+44.
(vii) F=E+H = 2E+4.
(viii) The width of the top row A+B = the width of the bottom row F+G+H = (E+28)+(E+44) = (E+8)+(E+4)+(2E+4); 2E+72=4E+16; 72-16=2E; E=28.
Width = A+B = 2E+72= 128
Height = A+C+G = (E+28)+(E+12)+(E+8) - 3E+48 = 132.
So area = 128*132 = 16896.
(i) H=E+4, consider its top edge which meet "4" and "E".
(ii) G=H+4=E+8, consider its Eastern edge which meet "4" and "H".
(iii) Similarly C=G+4=E+12.
(iv) The height of C and G is the same as D, E and H. (E+12)+(E+8)=D+E+(E+4). So D=16.
(v) A= C+D = (E+12)+16 = E+28.
(vi) B= A+C = (E+28)+16 = E+44.
(vii) F=E+H = 2E+4.
(viii) The width of the top row A+B = the width of the bottom row F+G+H = (E+28)+(E+44) = (E+8)+(E+4)+(2E+4); 2E+72=4E+16; 72-16=2E; E=28.
Width = A+B = 2E+72= 128
Height = A+C+G = (E+28)+(E+12)+(E+8) - 3E+48 = 132.
So area = 128*132 = 16896.
July 29th, 2020 at 10:37:56 AM
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Quote: Gialmere
...
Since this is an easy math puzzle, I did it an easy way.
I copied the picture using Microsoft's Snipping Tool and pasted into a new "drawing canvas" in a Word document.
I found that a square 0.1 inch on a side covered the picture of the little square. The other eight squares were 0.4, 0.7, 0.8, 0.9, 1.0, 1.4, 1.5, and 1.8 inches on a side. (Also, I checked that sums of adjacent groups of squares were consistent with these measurements.) My rectangle was 1.4 + 1.8 = 3.2 inches wide and 1.8 + 1.5 = 3.3 inches tall.
Therefore, the original rectangle is 40(3.2) = 128 cm wide and 40(3.3) = 132 cm tall for an area of (128)(132) = 16,896 cm2.
I copied the picture using Microsoft's Snipping Tool and pasted into a new "drawing canvas" in a Word document.
I found that a square 0.1 inch on a side covered the picture of the little square. The other eight squares were 0.4, 0.7, 0.8, 0.9, 1.0, 1.4, 1.5, and 1.8 inches on a side. (Also, I checked that sums of adjacent groups of squares were consistent with these measurements.) My rectangle was 1.4 + 1.8 = 3.2 inches wide and 1.8 + 1.5 = 3.3 inches tall.
Therefore, the original rectangle is 40(3.2) = 128 cm wide and 40(3.3) = 132 cm tall for an area of (128)(132) = 16,896 cm2.
July 29th, 2020 at 2:01:50 PM
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A Mersenne Prime is a prime number that is one less than a power of 2.
For example, 127 is a Mersenne prime, as 127 = 27 - 1, and 7 is prime.
Note that the power of 2 must also be prime, as 2 to the power of (a nonprime number > 1) - 1 cannot be prime.
In March of 2011, someone claimed to have found two Mersenne primes (prime numbers that are one less than powers of 2); one ended in the digits 785894583, and the other in 943626843.
Prove that neither of these numbers could possibly be a Mersenne prime.
For example, 127 is a Mersenne prime, as 127 = 27 - 1, and 7 is prime.
Note that the power of 2 must also be prime, as 2 to the power of (a nonprime number > 1) - 1 cannot be prime.
In March of 2011, someone claimed to have found two Mersenne primes (prime numbers that are one less than powers of 2); one ended in the digits 785894583, and the other in 943626843.
Prove that neither of these numbers could possibly be a Mersenne prime.
July 29th, 2020 at 2:09:35 PM
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Quote: ThatDonGuy...one ended in the digits 785894583, and the other in 943626843.
Prove that neither of these numbers could possibly be a Mersenne prime.
The last digits in powers of 2 go in a cycle of 4.
2^1 = 2
2^2 = 4
2^3 = 8
2^4 =16
2^5 =32
2^6 =64
Thus 2^(4n+1)=10x+2.
Thus 2^(4n+2)=10x+4.
Thus 2^(4n+3)=10x+8.
Thus 2^(4n+4)=10x+6.
Thus an odd power of 2 ends in either a 2 or 8.
So a mersenne prime ends in 1 or 7; i.e. not a 3,
2^1 = 2
2^2 = 4
2^3 = 8
2^4 =16
2^5 =32
2^6 =64
Thus 2^(4n+1)=10x+2.
Thus 2^(4n+2)=10x+4.
Thus 2^(4n+3)=10x+8.
Thus 2^(4n+4)=10x+6.
Thus an odd power of 2 ends in either a 2 or 8.
So a mersenne prime ends in 1 or 7; i.e. not a 3,
July 29th, 2020 at 2:46:09 PM
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Quote: charliepatrickThe last digits in powers of 2 go in a cycle of 4.
2^1 = 2
2^2 = 4
2^3 = 8
2^4 =16
2^5 =32
2^6 =64
Thus 2^(4n+1)=10x+2.
Thus 2^(4n+2)=10x+4.
Thus 2^(4n+3)=10x+8.
Thus 2^(4n+4)=10x+6.
Thus an odd power of 2 ends in either a 2 or 8.
So a mersenne prime ends in 1 or 7; i.e. not a 3,
July 29th, 2020 at 4:17:47 PM
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Quote: ThatDonGuyQuote: GialmereThe rectangle above is divided into 9 perfect squares.
The smallest square (colored black) has a side length of 4cm.
What is the area of the rectangle?
Label the eight squares as follows:
A - upper left
B - upper right
C - left edge, below A
D - touching A, B, C
E - below D
F - bottom left
G - bottom center
H - bottom right
G = E + 4
F = G + 4 = E + 8
C = F + 4 = E + 12
C + 4 = D + E -> E + 16 = D + E -> D = 16
A = C + D = E + 12 + 16 = E + 28
B = A + D = E + 44
B + D = E + H -> E + 44 + 16 = E + H -> H = 60
H = E + G = 2E + 4 -> E = 28
The eight squares A, B, C, D, E, F, G, H have side lengths 56, 72, 40, 16, 28, 36, 32, and 60, respectively
The width of the rectangle = A + B = 128
Check: it is also F + G + H = 128
The height of the rectangle = A + C + F = 132
Check: it is also B + H = 132
The area = 128 x 132 = 16,896
Quote: charliepatrickIf you call the squares starting at the top, A B, C D, E F, G H. Then if you start at the square NE of the small square (E). Then work clockwise.
(i) H=E+4, consider its top edge which meet "4" and "E".
(ii) G=H+4=E+8, consider its Eastern edge which meet "4" and "H".
(iii) Similarly C=G+4=E+12.
(iv) The height of C and G is the same as D, E and H. (E+12)+(E+8)=D+E+(E+4). So D=16.
(v) A= C+D = (E+12)+16 = E+28.
(vi) B= A+C = (E+28)+16 = E+44.
(vii) F=E+H = 2E+4.
(viii) The width of the top row A+B = the width of the bottom row F+G+H = (E+28)+(E+44) = (E+8)+(E+4)+(2E+4); 2E+72=4E+16; 72-16=2E; E=28.
Width = A+B = 2E+72= 128
Height = A+C+G = (E+28)+(E+12)+(E+8) - 3E+48 = 132.
So area = 128*132 = 16896.
Quote: ChesterDogSince this is an easy math puzzle, I did it an easy way.
I copied the picture using Microsoft's Snipping Tool and pasted into a new "drawing canvas" in a Word document.
I found that a square 0.1 inch on a side covered the picture of the little square. The other eight squares were 0.4, 0.7, 0.8, 0.9, 1.0, 1.4, 1.5, and 1.8 inches on a side. (Also, I checked that sums of adjacent groups of squares were consistent with these measurements.) My rectangle was 1.4 + 1.8 = 3.2 inches wide and 1.8 + 1.5 = 3.3 inches tall.
Therefore, the original rectangle is 40(3.2) = 128 cm wide and 40(3.3) = 132 cm tall for an area of (128)(132) = 16,896 cm2.
Correct! Including a Kobayashi Maru solve.
--------------------------------
What do you get when you cross two squares with an ogre?
A Shrektangle.
Have you tried 22 tonight? I said 22.
July 30th, 2020 at 8:00:02 AM
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/pic5559639.png)
A college student was studying for exams late at night when a power blackout occurred.
He lit two uniform candles of equal length but one thicker than the other. The thick candle could burn for four hours and the thin one an hour less. When he finally went to sleep, the thick candle was twice as long as the thin one.
How long did the student study by candle light?
Have you tried 22 tonight? I said 22.
July 30th, 2020 at 8:54:19 AM
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Let's assume the candles were originally 12" tall. So the large one (L) burns at 3" per hour, and the small one (S) at 4" per hour.
After 2 hours L has lost 6" so has 6" left; S has lost 8" so has 4" left. After 3 hours L has lost 9" so has 3" left, S has now just burnt out.
So let t be the time before 3 hours. The height of the large one is L=3+3t, the height of the small one is S=4t.
At the time of the solution L=2S.
So 8t=3+3t; 5t=3, t=3/5.
So after 2 2/5 hour...
...the large candle is 12-(2.4*3) = 12-7.2 = 4.8" tall
...the small candle is 12-(2.4*4) = 12-9.6 = 2.4" tall.
So the time is 2 hours 24 minutes.
After 2 hours L has lost 6" so has 6" left; S has lost 8" so has 4" left. After 3 hours L has lost 9" so has 3" left, S has now just burnt out.
So let t be the time before 3 hours. The height of the large one is L=3+3t, the height of the small one is S=4t.
At the time of the solution L=2S.
So 8t=3+3t; 5t=3, t=3/5.
So after 2 2/5 hour...
...the large candle is 12-(2.4*3) = 12-7.2 = 4.8" tall
...the small candle is 12-(2.4*4) = 12-9.6 = 2.4" tall.
So the time is 2 hours 24 minutes.
July 30th, 2020 at 10:59:41 AM
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I see I agree with Charlie, but I want to get credit anyway...
Let:
a = Time remaining on thick candle
b = Time remaining on thin candle
t = Time candle burned.
a = 4-t
b = 3-t
The height of each candle will be proportional to it's remaining time to it's beginning time.
Ratio of height left of a = (4-t)/4
Ratio of height left of b = (3-t)/3
Let's say each candle starts at 12". At some time x, the thick one will be twice as tall. In other words,
12*((4-t)/4) = 2*12*((3-t)/3)
3 * (4-t) = 8 * (3-t)
12 - 3t = 24 - 8t
5t = 12
t = 2.4
So, the candles have burned for 2.4 hours.
Good puzzle.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
July 30th, 2020 at 4:24:34 PM
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Can someone help me? I just want to post my own form questions but I don't know how.
Math is the only true form of knowledge
