Easy math puzzles

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?

Quote: CrystalMath

Show Spoiler

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: charliepatrick

yes

You 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, OB² = OA² + AB².
So AB²=OB²-OA².
So the area being painted = Pi AB².

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: 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?

Quote: ThatDonGuy

Here's what I get

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: charliepatrick

I agree with ThatDonGuy

Show Spoiler

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.

Quote: gordonm888

Here'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: Gialmere

Quote: gordonm888

Here'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.

An offhand guess

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

Quote: gordonm888

Here'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: ThatDonGuy

Quote: gordonm888

Here'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.

This is pretty much just a guess

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.

Quote: Wizard

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To 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.

Quote: gordonm888

Comments

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: Gialmere

...How much farther did your head travel than your feet?...

Quote: Gialmere

How much farther did your head travel than your feet?

Quote: charliepatrick

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The 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.
40 075.017	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: Wizard

Wiz answer

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: Gialmere

How many fingers do the aliens have?

Quote: Ace2

Agree 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

Quote: Wizard

Wiz answer

7.

Wiz solution

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: Ace2

Show Spoiler

Agree 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

Comic

Quote: ThatDonGuy

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?

Quote: Gialmere

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: Gialmere

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

Quote: Wizard

Wiz answer

Pr(CV & positive test)/Pr(positive test) =
0.001*1 / (.001*1 +0.999*0.05) =
0.001/0.0510 =
0.0196

Comic

Quote: Gialmere

White to move.

Can White mate in one?

Quote: ksdjdj

my attempt

At 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

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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

Yes

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).

Comic

Quote: rsactuary

answer

1888 and 2888 ?

Quote: Gialmere

What's the next element in this series:

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