## Poll

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10 votes (21.73%) |

**46 members have voted**

Quote:Gialmere

An amusement park wants to restore its Merry-Go-Round and hires a group of artists for the job.

One artist is assigned to paint the carousel floor which is formed by two concentric circles (an annulus). She wants to determine the area of the floor (shown in yellow in the figure below), so she will know how much paint to buy.

Because of all the machinery in the middle, she is unable to measure the radii of the two circles. However, she finds the length of a special chord to be 70 feet. This special chord is a chord of the larger circle and a tangent to the smaller circle. (See diagram below).

Can she determine the area of the Merry-Go-Round which she needs to paint using just that one measurement?

Surprisingly, yes.

Outer radius = R

Inner radius = r

R = sqrt(35^2 + r^2)

A = pi(R^2 - r^2) = pi*(35^2 + r^2 - r^2) = 1225pi

The area of the large circle is Pi r"

^{2}.

The area of the small circle is Pi r'

^{2}.

You want to know the difference.

r" = OB, r' = OA, OB

^{2}= OA

^{2}+ AB

^{2}.

So AB

^{2}=OB

^{2}-OA

^{2}.

So the area being painted = Pi AB

^{2}.

In this case AB = 35 ft so the area (using Pi=22/7 as one would be allowed to in the exams in the old days and is why the number is divisible by 7) (approx) = 35*35*22/7 = 3850 sq ft.

Quote:CrystalMath

Surprisingly, yes.

Outer radius = R

Inner radius = r

R = sqrt(35^2 + r^2)

A = pi(R^2 - r^2) = pi*(35^2 + r^2 - r^2) = 1225pi

Quote:charliepatrickYou use Pythagoras.

The area of the large circle is Pi r"^{2}.

The area of the small circle is Pi r'^{2}.

You want to know the difference.

r" = OB, r' = OA, OB^{2}= OA^{2}+ AB^{2}.

So AB^{2}=OB^{2}-OA^{2}.

So the area being painted = Pi AB^{2}.

In this case AB = 35 ft so the area (using Pi=22/7 as one would be allowed to in the exams in the old days and is why the number is divisible by 7) (approx) = 35*35*22/7 = 3850 sq ft.

Correct!

You have to admire the simple elegance of this one.

--------------------------------

I caught a cold on a carousel once.

There must have been something going around.

Once again you are messing around with an old computer Tic-Tac-Toe game.

This time you program it to randomly place an X on the grid, and then randomly place an O, and then an X and so on.

Played in this manner, what are the odds that...

...X will win the game?

...O will win the game?

...it will end in a draw?

Quote:Gialmere

Once again you are messing around with an old computer Tic-Tac-Toe game.

This time you program it to randomly place an X on the grid, and then randomly place an O, and then an X and so on.

Played in this manner, what are the odds that...

...X will win the game?

...O will win the game?

...it will end in a draw?

O wins: (.2880952...)

Draw: (.1269841...)

There is a coin shortage going on in the US and you happen to have plenty of change on hand. I ask you to give me change for $1.

How many different ways can you change an American dollar bill into coins?

The dollar can either be broken into coins of (i) 50c + 50c (ii) 50c + two quarters (iii) four quarters.

50c (not being 25c) can either be {50} or {10 10 10 10 10} = 2 ways = A B

25c can be {25} {10 10 5} {10 5 5 5} {5 5 5 5 5} {10 10 1s} {10 5 5 1s} (10 5 1s} {10 1s} {5 5 5 5 1s} {5 5 5 1s} {5 5 1s} {5 1s} {1s} = 13 ways = a b c d e f g h i j k l m

(i) 50+50 can be AA,BB,AB = 3 ways

(ii) 50+25+25 can be (A or B) 2 ways, with (pair e.g. aa 13 ways) or (dispair e.g. ab) 13*12/2 = 78 ways; = 2 * (13+78) = 182 ways

(iii) 25+25+25+25 can be any permutations of four of (a...m)

(A) Four same (aaaa) = 13 ways

(B) Three/One (aaab) = 13*12 = 156 ways

(C) Two/Two (aabb) = 13*12/2 = 78 ways

(D) Two/One/One (aabc) = 13* 12*11/2 = 858 ways

(E) One/One/One/One (abcd) = 13*12*11*10/4/3/2/1 = 715 ways

Total = 13+156+78+858+715 = 1820 ways

So grand total = 3 + 182 + 1820 = 2005 ways.

If you use only nickels and pennies, there are 21 ways: 100 pennies, 95 pennies + 1 nickel, 90 pennies + 2 nickels, ..., 5 pennies + 19 nickels, 20 nickels.

Each dime added removes two of the methods; you can have up to 10 dimes, so there are 21 + 19 + 17 + ... + 1 = 121 ways with just pennies, nickels, and dimes.

With one quarter, you can have:

no dimes and 0, 1, 2, ..., 15 nickels

1 dime and 0, 1, ..., 13 nickels

2 dimes and 0, 1, ..., 11 nickels

3 dimes and 0, 1, ..., 9 nickels

4 dimes and 0, 1, ..., 7 nickels

5 dimes and 0, 1, ..., 5 nickels

6 dimes and 0, 1, 2, 3 nickels

7 dimes and 0 or 1 nickels

There are 16 + 14 + 12 + ... + 2 = 72 ways with one quarter

With two quarters, you can have:

no dimes and 0, 1, 2, ..., 10 nickels

1 dime and 0, 1, 2, ..., 8 nickels

2 dimes and 0, 1, 2, ..., 6 nickels

3 dimes and 0, 1, 2, 3, 4 nickels

4 dimes and 0, 1, 2 nickels

5 dimes

There are 11 + 9 + 7 + 5 + 3 + 1 = 36 ways with two quarters

For each way with two quarters, replace the two quarters with a half dollar; there are 36 ways of this

With three quarters, you can have:

no dimes and 0, 1, 2, 3, 4, 5 nickels

1 dime and 0, 1, 2, 3 nickels

2 dimes and 0 or 1 nickels

There are 6 + 4 + 2 = 12 ways with three quarters

For each way, replace two of the quarters with a half dollar; this is another 12 ways

Finally, there are:

Four quarters

One half dollar, two quarters

Two half dollars

One dollar coin

The total is 121 + 72 + 36 + 36 + 12 + 12 + 4 = 293

Generating functions are not my forte.

(I) $; 50+50; 50+25+25; 25+25+25+25 = 4 Ways.

(II) else use 25c+ to use up None: 0, One: 25, Two: 50/25 25, Three: 50 25/25 25 25 quarters' worth.

(a) No quarters could be

10x10c (no 5c) - 1 way

9x10c (2x5c) - (could be 2 0, 1 5, 0 10) - 3 ways . . .

0x10c (20x5c) - (could be 20 0, 19 5, ... ,0 100.) - 21 ways

1+3+...+19+21 = 121 Ways.

(b) One quarter could be

7x10c (1 5c) - 2 ways . . .

0x10c (15 5c) - 16 ways

2+4+...+14+16 = 72 Ways.

(c) Two quarters (or 1 50c) could be

5x10c (no 5c) - 1 way . . .

0x10c (10x5c) - 11 ways

1+3+5+7+9+11 = 36 : then there are two ways to be here so 72 Ways.

(d) Three quarters (or 50c 25c) could be

2x10c (1 5c) - 2 ways . . .

0x10c (5 5c) - 6 ways

2+4+6 = 12 : then there are two ways to be here so 24 Ways.

Ways: 4 121 72 72 24 = 293 WAYS.

This is in a European language:

Uspostaviti denominacije kovanica takve da zahtijeva minimalni broj kovanica danih dvoje ljudi tako da jedan može kupiti od drugog bilo što u rasponu od jedne do tisuću novčića

Answers in spoiler boxes please.

Quote:ThatDonGuy

If you use only nickels and pennies, there are 21 ways: 100 pennies, 95 pennies + 1 nickel, 90 pennies + 2 nickels, ..., 5 pennies + 19 nickels, 20 nickels.

Each dime added removes two of the methods; you can have up to 10 dimes, so there are 21 + 19 + 17 + ... + 1 = 121 ways with just pennies, nickels, and dimes.

With one quarter, you can have:

no dimes and 0, 1, 2, ..., 15 nickels

1 dime and 0, 1, ..., 13 nickels

2 dimes and 0, 1, ..., 11 nickels

3 dimes and 0, 1, ..., 9 nickels

4 dimes and 0, 1, ..., 7 nickels

5 dimes and 0, 1, ..., 5 nickels

6 dimes and 0, 1, 2, 3 nickels

7 dimes and 0 or 1 nickels

There are 16 + 14 + 12 + ... + 2 = 72 ways with one quarter

With two quarters, you can have:

no dimes and 0, 1, 2, ..., 10 nickels

1 dime and 0, 1, 2, ..., 8 nickels

2 dimes and 0, 1, 2, ..., 6 nickels

3 dimes and 0, 1, 2, 3, 4 nickels

4 dimes and 0, 1, 2 nickels

5 dimes

There are 11 + 9 + 7 + 5 + 3 + 1 = 36 ways with two quarters

For each way with two quarters, replace the two quarters with a half dollar; there are 36 ways of this

With three quarters, you can have:

no dimes and 0, 1, 2, 3, 4, 5 nickels

1 dime and 0, 1, 2, 3 nickels

2 dimes and 0 or 1 nickels

There are 6 + 4 + 2 = 12 ways with three quarters

For each way, replace two of the quarters with a half dollar; this is another 12 ways

Finally, there are:

Four quarters

One half dollar, two quarters

Two half dollars

One dollar coin

The total is 121 + 72 + 36 + 36 + 12 + 12 + 4 = 293

Quote:charliepatrickI agree with ThatDonGuy

You can either use just 25c or higher, else get 10c and lower involved.

(I) $; 50+50; 50+25+25; 25+25+25+25 = 4 Ways.

(II) else use 25c+ to use up None: 0, One: 25, Two: 50/25 25, Three: 50 25/25 25 25 quarters' worth.

(a) No quarters could be

10x10c (no 5c) - 1 way

9x10c (2x5c) - (could be 2 0, 1 5, 0 10) - 3 ways . . .

0x10c (20x5c) - (could be 20 0, 19 5, ... ,0 100.) - 21 ways

1+3+...+19+21 = 121 Ways.

(b) One quarter could be

7x10c (1 5c) - 2 ways . . .

0x10c (15 5c) - 16 ways

2+4+...+14+16 = 72 Ways.

(c) Two quarters (or 1 50c) could be

5x10c (no 5c) - 1 way . . .

0x10c (10x5c) - 11 ways

1+3+5+7+9+11 = 36 : then there are two ways to be here so 72 Ways.

(d) Three quarters (or 50c 25c) could be

2x10c (1 5c) - 2 ways . . .

0x10c (5 5c) - 6 ways

2+4+6 = 12 : then there are two ways to be here so 24 Ways.

Ways: 4 121 72 72 24 = 293 WAYS.

Correct!

------------------------------

I heard that the Treasury Department wants to get rid of the penny and I'm not sure how to feel about it...

On one hand I'm opposed to change but, on the other hand, I'm opposed to change.

Quote:gordonm888Here's a puzzle with the same theme, but it has some 'twists' to it:

This is in a European language:

Uspostaviti denominacije kovanica takve da zahtijeva minimalni broj kovanica danih dvoje ljudi tako da jedan može kupiti od drugog bilo što u rasponu od jedne do tisuću novčića

Answers in spoiler boxes please.

Quote:GialmereQuote:gordonm888Here's a puzzle with the same theme, but it has some 'twists' to it:

This is in a European language:

Uspostaviti denominacije kovanica takve da zahtijeva minimalni broj kovanica danih dvoje ljudi tako da jedan može kupiti od drugog bilo što u rasponu od jedne do tisuću novčića

Answers in spoiler boxes please.I'll just guess casino check denominations: 1c, 5c, 25c, 100c, 500c, 1000c. I'm sure they've thought about this..

Congrats for getting through the first two hurdles and understanding the question. But this is not the correct answer.

So for one person to pay the sum to the other 1 thru 1000 in coins you would need 1 2 2 and 199 5's. Total = 202 coins.

For one person to pay the sum to the other where the other can have initial coin(s) is still 204 as the recipient could have a small coin and be able to give change, but the sender would need another 5 (eg 1 2 200 5's, 2).

For either person to pay the other would need (1 2 200 5's) opposite (200 5's 1) (or other similar holdings e.g.: 12-2, 1 2 2 199 5's - 200 5's). Total = 402 coins.

If you were allowed banknotes as well, then the sender needs something like 1 2 2 5 10 10 20 50 100 100 200 500 and, if two way, as the sender can send back 1-1000 the receiver only needs 1000, so can just have 500 500. If 500's aren't available then use 200's instead.

Quote:gordonm888Here's a puzzle with the same theme, but it has some 'twists' to it:

This is in a European language:

Uspostaviti denominacije kovanica takve da zahtijeva minimalni broj kovanica danih dvoje ljudi tako da jedan može kupiti od drugog bilo što u rasponu od jedne do tisuću novčića

Answers in spoiler boxes please.

I can't come up with a better solution than 10 denominations - 1, 2, 4, 8, 16, 32, 64, 128, 256, and 512. In this case, the buyer has one of each coin, and the seller needs none as the buyer can always make exact change, so the number of coins in total is 10.

My first thought was, the problem was similar to a "how do you weigh anything from 1 to 1000 with as few weights as possible" problem, and the coins would be 1, 3, 9, 27, 81, 243, and 729 (for example, to buy something worth 100, the buyer would pay an 81, a 27, and a 1, and the seller would give back a 9 in change) , but it turns out that the seller also needs coins of 1, 3, 9, 27, 81, and 243, so that's 13 coins in all. The difference is, in the weights problem, you only need one weight of 9 (you place the 81, 27, and 1 weights on one side of the balance, and the 100-weight item and the 9 weight on the other), while in the spending problem, both the buyer and the seller need separate 9-value coins.

Let a(x) = number of ways to make x cents, using only pennies and nickles, where x is evening divisible by 5.

Let y = x/5

a(x) = 1+y

Let b(x) = number of ways to make x cents, using only pennies, nickles, and dimes, where x is evening divisible by 5.

b(0)=1

b(5)=2

b(x) = a(x) + b(x-10), where x>=10.

Let c(x) = number of ways to make x cents, using only pennies, nickles, dimes, and quarters where x is evening divisible by 25.

c(0) = 1

c(x) = b(x) + c(x-25), where x>=25.

Let d(x) = number of ways to make x cents, using only pennies, nickles, dimes, quarters, and half dollars where x is evening divisible by 50.

d(0) = 1

d(x) = c(x) + d(x-50), where x>=50.

x | a(x) | b(x) | c(x) | d(x) |
---|---|---|---|---|

0 | 1 | 1 | 1 | |

5 | 2 | 2 | ||

10 | 3 | 4 | ||

15 | 4 | 6 | ||

20 | 5 | 9 | ||

25 | 6 | 12 | 13 | |

30 | 7 | 16 | ||

35 | 8 | 20 | ||

40 | 9 | 25 | ||

45 | 10 | 30 | ||

50 | 11 | 36 | 49 | 50 |

55 | 12 | 42 | ||

60 | 13 | 49 | ||

65 | 14 | 56 | ||

70 | 15 | 64 | ||

75 | 16 | 72 | 121 | |

80 | 17 | 81 | ||

85 | 18 | 90 | ||

90 | 19 | 100 | ||

95 | 20 | 110 | ||

100 | 21 | 121 | 242 | 292 |

So, the answer is 292, plus 1 for the $1 coin, equals 293.

Quote:ThatDonGuyQuote:gordonm888

This is in a European language:

Uspostaviti denominacije kovanica takve da zahtijeva minimalni broj kovanica danih dvoje ljudi tako da jedan može kupiti od drugog bilo što u rasponu od jedne do tisuću novčića

Answers in spoiler boxes please.

I can't come up with a better solution than 10 denominations - 1, 2, 4, 8, 16, 32, 64, 128, 256, and 512. In this case, the buyer has one of each coin, and the seller needs none as the buyer can always make exact change, so the number of coins in total is 10.

My first thought was, the problem was similar to a "how do you weigh anything from 1 to 1000 with as few weights as possible" problem, and the coins would be 1, 3, 9, 27, 81, 243, and 729 (for example, to buy something worth 100, the buyer would pay an 81, a 27, and a 1, and the seller would give back a 9 in change) , but it turns out that the seller also needs coins of 1, 3, 9, 27, 81, and 243, so that's 13 coins in all. The difference is, in the weights problem, you only need one weight of 9 (you place the 81, 27, and 1 weights on one side of the balance, and the 100-weight item and the 9 weight on the other), while in the spending problem, both the buyer and the seller need separate 9-value coins.

CORRECT!

I laid a trap with the mathematical question -about using the least coins of whatever denominations you care to define so as to allow the buyer and sellor to transact anything from one penny to 1000 pennies. I thought,, as ThatDonGuy said, that people would give an answer of 13 or 14 coins that had denominations of multiples of 3. But ThatDonGuy got it correct: just 10 coins of 2^n denominations.

If I had asked you to minimize denominations, then 7 denominations of 1, 3, 9, 27, 81, 243 and 729 would have been the correct answer.

Quote:WizardTo return to the question on change for a dollar, here is my answer.

Let a(x) = number of ways to make x cents, using only pennies and nickles, where x is evening divisible by 5.

Let y = x/5

a(x) = 1+y

Let b(x) = number of ways to make x cents, using only pennies, nickles, and dimes, where x is evening divisible by 5.

b(0)=1

b(5)=2

b(x) = a(x) + b(x-10), where x>=10.

Let c(x) = number of ways to make x cents, using only pennies, nickles, dimes, and quarters where x is evening divisible by 25.

c(0) = 1

c(x) = b(x) + c(x-25), where x>=25.

Let d(x) = number of ways to make x cents, using only pennies, nickles, dimes, quarters, and half dollars where x is evening divisible by 50.

d(0) = 1

d(x) = c(x) + d(x-50), where x>=50.

x a(x) b(x) c(x) d(x) 0 1 1 1 5 2 2 10 3 4 15 4 6 20 5 9 25 6 12 13 30 7 16 35 8 20 40 9 25 45 10 30 50 11 36 49 50 55 12 42 60 13 49 65 14 56 70 15 64 75 16 72 121 80 17 81 85 18 90 90 19 100 95 20 110 100 21 121 242 292

So, the answer is 292, plus 1 for the $1 coin, equals 293.

Correct!

Quote:gordonm888If I had asked you to minimize denominations, then 7 denominations of 1, 3, 9, 27, 81, 243 and 729 would have been the correct answer.

If you want to minimize the number of denominations, then 1 is the answer - you only need 1 demonimation; 1000 coins of value 1 each.

You are a six foot tall athlete who has just (somehow) walked entirely around the earth along the equator and returned to your starting point.

How much farther did your head travel than your feet?

Quote:Gialmere...How much farther did your head travel than your feet?...

That was easy , but , a person doesn't walk that way!!

1) The head actually bobs up and down as you walk.

Let's assume the legs are 3ft long and the torso/head are the other 3ft. Also assume the legs form an equilateral triangle at the start of each step (makes the maths easier) (if in doubt think how soccer referees pace out ten yards).

For each step one foot will stay put, while the other foot will move from being 3ft behind to 3ft ahead. So, on average, each foot travels 3ft per step. (We'll ignore the fact you have two feet and thoughts that the feet, together, travel twice as far as your head as there's two of them and only one head!)

Meanwhile your head follows a similar path to the top of the moving leg, which is a circle of radius 3ft for 60 degrees. The total circumference of a 3ft circle is 2 Pi R = 6 Pi, so your head travels 60 degrees which is Pi.

Therefore for each 3ft your feet travel, your head travels Pi feet.

According to https://web.archive.org/web/20110424104419/http://home.online.no/~sigurdhu/WGS84_Eng.html the radius at the equator is 40 075.017 km (and I used to think it was 40,000Km).

So if your feet travel that far, your head (ignoring the effect of its height) = circumference of equator * (Pi-3).

2) The head is still farther away from the earth than your feet, so the circumference of your head should be lbigger due to your average height as you walk.

The average height of the head above the ground can be calculated by looking at the area under the curve of the arc of 60 degrees.

This consists of a 1/6 of a circle together with two triangles (half equilateral triangles with sides 3ft long), then a 3x3 square.

Extra distance of head due to walking. | |
---|---|

40 075.017 | Km = circumference |

131 479 714.567 | Ft (using inches/meters 39.370078740157) |

18 616 561.679 | ^ times (Pi-3) |

18 616 561ft 8in | Difference |

Extra distance of head due to average height. | |

2.598 076 | Height of triangles = SQRT (9-2.25) |

1.500 000 | Width of each triangle |

3.897 114 | Area of the two triangles |

4.712 389 | Area of arc of circle (Pi R^2/6) |

9.000 000 | Area of square (3x3) |

17.609 503 | Total area |

5.869 834 | Average height of head above ground |

36.881 257 | Extra distance due to height (2 Pi R) |

36ft 10in | Distance |

18 616 598ft 6in | Total extra distance |

Quote:GialmereHow much farther did your head travel than your feet?

12*pi feet.

Solution:

Let r be the radius of the earth. The extra distance covered is.

2*pi*(r+6) - 2*pi*r = 12*pi

Quote:charliepatrickThe simple answer is the feet go round a circle of radius R and the head at R+6. The circumference of a circle is 2 Pi R, so the difference is 2 Pi ((R+6)-R) = 12 Pi.

That was easy , but , a person doesn't walk that way!!

1) The head actually bobs up and down as you walk.

Let's assume the legs are 3ft long and the torso/head are the other 3ft. Also assume the legs form an equilateral triangle at the start of each step (makes the maths easier) (if in doubt think how soccer referees pace out ten yards).

For each step one foot will stay put, while the other foot will move from being 3ft behind to 3ft ahead. So, on average, each foot travels 3ft per step. (We'll ignore the fact you have two feet and thoughts that the feet, together, travel twice as far as your head as there's two of them and only one head!)

Meanwhile your head follows a similar path to the top of the moving leg, which is a circle of radius 3ft for 60 degrees. The total circumference of a 3ft circle is 2 Pi R = 6 Pi, so your head travels 60 degrees which is Pi.

Therefore for each 3ft your feet travel, your head travels Pi feet.

According to https://web.archive.org/web/20110424104419/http://home.online.no/~sigurdhu/WGS84_Eng.html the radius at the equator is 40 075.017 km (and I used to think it was 40,000Km).

So if your feet travel that far, your head (ignoring the effect of its height) = circumference of equator * (Pi-3).

2) The head is still farther away from the earth than your feet, so the circumference of your head should be lbigger due to your average height as you walk.

The average height of the head above the ground can be calculated by looking at the area under the curve of the arc of 60 degrees.

This consists of a 1/6 of a circle together with two triangles (half equilateral triangles with sides 3ft long), then a 3x3 square.

Extra distance of head due to walking. 40 075.017Km = circumference 131 479 714.567Ft (using inches/meters 39.370078740157) 18 616 561.679^ times (Pi-3) 18 616 561ft 8inDifference Extra distance of head due to average height. 2.598 076Height of triangles = SQRT (9-2.25) 1.500 000Width of each triangle 3.897 114Area of the two triangles 4.712 389Area of arc of circle (Pi R^2/6) 9.000 000Area of square (3x3) 17.609 503Total area 5.869 834Average height of head above ground 36.881 257Extra distance due to height (2 Pi R) 36ft 10inDistance 18 616 598ft 6inTotal extra distance

Very impressive CP!

Quote:Wizard

12*pi feet.

Solution:

Let r be the radius of the earth. The extra distance covered is.

2*pi*(r+6) - 2*pi*r = 12*pi

Correct!

------------------------------

Fun Fact: If the human population all held hands around the equator a significant portion of them would drown.

Your feet must travel a distance = 2* π * 3963.2 miles x 5280 feet/mile= 2* π * 20,595,696 feet

Assuming the center of you head is 5.8 feet above the ground your head must travel 2* &pi * (20,595,696 +5.8) feet.

So, the difference is 2*&pi *5.8 ft = 36.44 feet approximately.

I was writing this post while gialmere posted the above.

You are an astronomer who has discovered intelligent life on a distant planet. Such is the resolution of your super telescope, you can see right into what appears to be a mathematics classroom. On the board the following equations are written.

13 + 15 = 31

10 x 10 = 100

6 x 3 = 24

How many fingers do the aliens have?

7.

I needed only the first equation:

Let x be the answer.

(t+3) + (t+5) = (3t + 1)

2t + 8 = 3t +1

t = 7

The second equation will be true for any number of fingers.

If t=7, then 6*3 = 24.

Quote:GialmereHow many fingers do the aliens have?

It may depend on whether or not 0, 1, 2, ... mean the same thing to the aliens as they do on Earth.

If you saw:

AC + AE = CA

AO x AO = AOO

F x C = BD

would there be any other possible solutions?

Let b be the base; assume b is an integer > 1

AO x AO = AOO means (Ab + O)(Ab + O) = A b^2 + O b + O

If A > 1, then either the first digit of the product > A or the product has more than three digits

Even if the aliens did use leading zeroes, if A = 0, then equation becomes O^2 = Ob + O, or O = b + 1, but this is impossible as this would make O a 2-digit number in base b

Therefore, A = 1

The equation is now 1O x 1O = 1OO

b^2 + 2O b + O^2 = b^2 + O b + O

Either O = 0, or 2 b + O = b + 1, but 2 b + O = b + 1 means b = 1 - O

Therefore, O = 0

I only used the last equation, which immediately tells you the base must be at least 6.

Six times three is eighteen. (18 - 4) / 2 = 7

I think the middle equation would be true for any base

Quote:Ace2Agree with Wizard.

I only used the last equation, which immediately tells you the base must be at least 6.

Six times three is eighteen. (18 - 4) / 2 = 7

I think the middle equation would be true for any base

next time, please use spoilers

Quote:Wizard

7.

I needed only the first equation:

Let x be the answer.

(t+3) + (t+5) = (3t + 1)

2t + 8 = 3t +1

t = 7

The second equation will be true for any number of fingers.

If t=7, then 6*3 = 24.

Quote:Ace2Agree with Wizard.

I only used the last equation, which immediately tells you the base must be at least 6.

Six times three is eighteen. (18 - 4) / 2 = 7

I think the middle equation would be true for any base

Correct!

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Quote:ThatDonGuyIt may depend on whether or not 0, 1, 2, ... mean the same thing to the aliens as they do on Earth.

If you saw:

AC + AE = CA

AO x AO = AOO

F x C = BD

would there be any other possible solutions?

It turns out there would be an infinite number of solutions with O = 0, A = 1, and B = 2

As shown in an earlier post, O = 0 and A = 1

Let b be the base; note b >= 7 as there are (at least) seven different digits

For 1C + 1E = C1, there is a carry, so the second digit from the right in the answer is 3 (1 + 1 + the carried 1); C = 3, and E = b - 2

The last equation becomes F x 3 = BD

If B = 2, F = b - 1, and D = b - 3, the last equation becomes:

3 (b - 1) = 2 b + (b - 3) = 3b - 3

which is always true

Example: in base 8, D = 5, E = 6, and F = 7

13 + 16 = 31

10 x 10 = 100

7 x 3 = 25

Uh oh. The coronavirus has mutated and the new strain is deadly. Anyone who contracts this new, killer Covid will die within a week.

The good (if you can call it that) news is that only 0.1% of the population will catch it. The bad news is you just tested positive for it.

Trying to keep your voice calm, you ask the doctor who administered the test how accurate it is. She tells you that it's 95% accurate, that is 5% of the people will get a false positive. (Assume there are no false negatives.)

What is the probability that you now have an appointment with the grim reaper?

Quote:GialmereUh oh. The coronavirus has mutated and the new strain is deadly. Anyone who contracts this new, killer Covid will die within a week.

The good (if you can call it that) news is that only 0.1% of the population will catch it. The bad news is you just tested positive for it.

Trying to keep your voice calm, you ask the doctor who administered the test how accurate it is. She tells you that it's 95% accurate, that is 5% of the people will get a false positive. (Assume there are no false negatives.)

What is the probability that you now have an appointment with the grim reaper?

Assuming that "only 0.1% of the population will catch it" means "in any random selection of 1000 people, one person will ever have the disease":

P(true positive) = 1/1000 x 19/20 = 19 / 20,000

P(false positive) = 999/1000 x 1/20 = 999 / 20,000

P(I have the disease) = 19 / (19 + 999) = 19 / 1018

However, the initial assumption is false.

No false negatives

Pr(CV & positive test)/Pr(positive test) =

0.001*1 / (.001*1 +0.999*0.05) =

0.001/0.0510 =

0.0196

Quote:Gialmere

What is the probability that you now have an appointment with the grim reaper?

We all have an appointment with GR, don't we? Maybe not on our calendars, but certainly on his! ;)

Quote:Wizard

Pr(CV & positive test)/Pr(positive test) =

0.001*1 / (.001*1 +0.999*0.05) =

0.001/0.0510 =

0.0196

Correct!

If you test 1,000 people you expect 51 positives. 50 will be false positives and 1 will be a carrier. So you enter the lottery from hell with a 1 in 51 chance of being a dead man walking.

= 0.0196078431...

Still, surprisingly good odds considering.

--------------------------

White to move.

Can White mate in one?

Even though I could have a closer look later, currently I don't see any other move that could be made to "(check) mate in one", so my answer would be no (assuming a "normal north/south" board orientation).

Quote:Gialmere

White to move.

Can White mate in one?

Yes.

I don't know if White started on the left or right side of the picture, but if it started on the left, a White pawn can advance to be promoted to a queen for mate. And likewise, if White started on the right side of the picture, White's other pawn could do the same.

(Of course, White didn't start in the foreground or the distant side because the square on the near right is black.)

According to the rules of chess, each player's lower-left corner is black. This means the game is being played left-to-right.

Regardless of which side of the board is white, the winning move is the same: f8=Q (i.e. move the far pawn, then promote to queen).

Quote:ksdjdjAt first glance I can't see any move when looking "north/south" (the "traditional/normal way of viewing the board"). But if the game is being played on an "east/west" board, then you could possibly convert the pawn to a queen and mate in one that way (the pawn nearest to the "white-square bishop").

Even though I could have a closer look later, currently I don't see any other move that could be made to "(check) mate in one", so my answer would be no (assuming a "normal north/south" board orientation).

Quote:ChesterDog

Yes.

I don't know if White started on the left or right side of the picture, but if it started on the left, a White pawn can advance to be promoted to a queen for mate. And likewise, if White started on the right side of the picture, White's other pawn could do the same.

(Of course, White didn't start in the foreground or the distant side because the square on the near right is black.)

Quote:ThatDonGuy

According to the rules of chess, each player's lower-left corner is black. This means the game is being played left-to-right.

Regardless of which side of the board is white, the winning move is the same: f8=Q (i.e. move the far pawn, then promote to queen).

Correct!

---------------------------------

Starting with the year AD 1, which year in history has had the greatest number of Roman Numerals in it?

And which year will it be when this record is surpassed?

Quote:rsactuary1888 and 2888 ?

Correct!

MDCCCLXXXVIII

In 2888, there will be 14 characters:

MMDCCCLXXXVIII

---------------------------------

There are eleven types of people in the world...

...those that understand Roman numerals and those that don’t.

What's the next element in this series:

1, 11, 21, 1211, 111221, ... ?

Quote:Gialmere

What's the next element in this series:

1, 11, 21, 1211, 111221, ... ?

This I've seen before--it's 312211, but I won't explain it.