Easy math puzzles

Quote: ThatDonGuy

Dice where 2s and 5s "don't count" remind me of Scarney Dice (created by John Scarne), where the 2s and 5s actually say "Dead".

Answer

Hopefully I avoided any idiot math errors this time; the number appears to be confirmed by simulation.

Let E(n) be the expected score with N dice remaining
E(0) = 0

E(1) = 2/3 (E(1) + 7/2)
= 7

E(2) = 4/9 (E(2) + 7) + 4/9 E(1)
= 4/9 E(2) + 56/9
= 56/5

E(3) = 8/27 (E(3) + 21/2) + 12/27 E(2) + 6/27 E(1)
= 8/27 E(3) + 84/27 + 12/27 x 56/5 + 6/27 x 7
19 E(3) = 84 + 12 x 56/5 + 42
E(3) = (126 + 12 x 56/5) / 19
= (126 x 5 + 12 x 56) / 95
= 1302 / 95

E(4) = 16/81 (E(4) + 14) + 32/81 E(3) + 24/81 E(2) + 8/81 E(1)
= 16/81 E(4) + 224/81 + 32/81 x 1302/95 + 24/81 x 56/5 + 8/81 x 7
65 E(4) = 224 + 32 x 1302 / 95 + 24 x 56 / 5 + 56
E(4) = (280 x 475 + 41,664 x 5 + 1344 x 95) / (95 x 5 x 65)
= (280 x 95 + 41,664 + 1344 x 19) / (95 x 65)
= 3752 / 247

E(5) = 32/243 (E(5) + 35/2) + 80/243 E(4) + 80/243 E(3) + 40/243 E(2) + 10/243 E(1)
243 E(5) = 32 E(5) + 560 + 80 x 3752 / (19 x 13) + 80 x 1302 / (19 x 5) + 448 + 70
211 E(5) = 1078 + 80 x 3752 / (19 x 13) + 80 x 1302 / (19 x 5)
(211 x 19 x 65) E(5) = 1078 x 19 x 65 + 80 x 3752 x 5 + 80 x 1302 x 13
E(5) = 837,242 / 52,117, or about 16.065

Quote: Gialmere

...
On his TV show in 2007, comedian David Letterman made the following quip...

"In 1626, we bought Manhattan from the Indians for $24. Today, it would be worth 730 trillion dollars."

At what interest rate would you have to invest $24 in 1626 so that it would be worth $730 trillion in 2007?

Assume that the rate stays constant and that it is compounded annually.

Give your answer as a percent correct to two decimal places (e.g., like 3.74%).

Quote: Gialmere

On his TV show in 2007, comedian David Letterman made the following quip...

"In 1626, we bought Manhattan from the Indians for $24. Today, it would be worth 730 trillion dollars."

Quote: Joeman

I get...

8.49%

For annual compounding...
Today's value = Initial principal * (1 + rate) ^ years

--or--

$730 Trillion = $24(1+rate) ^ 381

Dividing both sides by the initial principal and then taking the 381st root, you get:

1 + rate = 1.0849

So the rate = 0.0849 or 8.49%

Show Spoiler

Quote: unJon

I get the other answer. Think you subtracted wrong. 394 years not 381.

Quote: Wizard

At the risk of making a political comment, that makes the national debt seem not so bad.

Quote: Joeman

I read the question as asking about the value in 2007 when the statement was made, not today's value.

Quote: Joeman

If the US were to sell Manhattan at that valuation, the government could pay off all its debts and have enough left over to make each US citizen a millionaire twice over!

Quote: gordonm888

...weird Texas hold-em poker...The deck is 10 to Ace, 4 suits; 20 cards.

1. What is the lowest possible 5-card poker hand that a player can make with his two hole cards and five common cards?

2. Given that hand categories are: Royal Flush, Straight Flush, Quads, etc., which hand category is the most probable and with what probability?

Quote: charliepatrick

Quote: gordonm888

...weird Texas hold-em poker...The deck is 10 to Ace, 4 suits; 20 cards.

1. What is the lowest possible 5-card poker hand that a player can make with his two hole cards and five common cards?

2. Given that hand categories are: Royal Flush, Straight Flush, Quads, etc., which hand category is the most probable and with what probability?

Show Spoiler

The lowest hand is Two Pairs, because if before the river you only had one pair then you would have all five ranks making a straight. Similarly Trips makes a straight.

RF	420
Quads	2800
FH	29280
Straight	27740
Two Pairs	17280

So Full House is most likely 29280/77520 ~= 37.77%

Quote: gordonm888

Quote: charliepatrick

Quote: gordonm888

...weird Texas hold-em poker...The deck is 10 to Ace, 4 suits; 20 cards.

1. What is the lowest possible 5-card poker hand that a player can make with his two hole cards and five common cards?

2. Given that hand categories are: Royal Flush, Straight Flush, Quads, etc., which hand category is the most probable and with what probability?

Show Spoiler

The lowest hand is Two Pairs, because if before the river you only had one pair then you would have all five ranks making a straight. Similarly Trips makes a straight.

RF	420
Quads	2800
FH	29280
Straight	27740
Two Pairs	17280

So Full House is most likely 29280/77520 ~= 37.77%

For question #1, your answer was a hand category and not the lowest possible hand. In normal Texas holdem, the lowest possible 5 card poker hand is 9-8-7-5-4, not "No Pair."

For question#2, I calculate a different number.

Quote: Joeman

I get...

8.49%

For annual compounding...
Today's value = Initial principal * (1 + rate) ^ years

--or--

$730 Trillion = $24(1+rate) ^ 381

Dividing both sides by the initial principal and then taking the 381st root, you get:

1 + rate = 1.0849

So the rate = 0.0849 or 8.49%

Show Spoiler

Quote: Joeman

The lowest hand would have to be

Two Pair: Queens & Jacks with a King kicker. The combination of hand + board would be: T, T, J, J, Q, Q, K

If there are only two pair among the 7 cards, there must also be a straight. So, to get a 5-card hand of two pair, there must be three pair among the 7 cards.

Quote: Joeman

The lowest hand would have to be

Two Pair: Queens & Jacks with a King kicker. The combination of hand + board would be: T, T, J, J, Q, Q, K

If there are only two pair among the 7 cards, there must also be a straight. So, to get a 5-card hand of two pair, there must be three pair among the 7 cards.

Quote: gordonm888

...For question#2, I calculate a different number.

Quote: charliepatrick

Show Spoiler

I've tried a different method as the distribution of cards leads to a unique category (except Straights may be SFs and there are 420 of those 4*15*14/2).

C1	C2	C3	C4	C5	Combins					Ways				Quads	FullHouse	Straight/SF	TwoPairs
4	3	0	0	0	1	4	1	1	1	5	4	1	80	80	0	0	0
4	2	1	0	0	1	6	4	1	1	5	4	3	1 440	1 440	0	0	0
4	1	1	1	0	1	4	4	4	1	5	4	1	1 280	1 280	0	0	0
3	3	1	0	0	4	4	4	1	1	10	3	1	1 920	0	1 920	0	0
3	2	2	0	0	4	6	6	1	1	5	6	1	4 320	0	4 320	0	0
3	2	1	1	0	4	6	4	4	1	5	4	3	23 040	0	23 040	0	0
3	1	1	1	1	4	4	4	4	4	5	1	1	5 120	0	0	5 120	0
2	2	2	1	0	6	6	6	4	1	10	2	1	17 280	0	0	0	17 280
2	2	1	1	1	6	6	4	4	4	10	1	1	23 040	0	0	23 040	0
													77 520	2 800	29 280	28 160	17 280

Quote: Gialmere

...can you fill in the empty boxes...

P	W	L	T	F	A	Points
4	2	1	1	7	1	5
2	0	2	0	2	6	0
4	1	2	1	5	10	3
3	0	0	3	4	4	3
3	2	0	1	5	2	5

Quote: charliepatrick

Show Spoiler

(a) C Played 3, 3 pts with no losses -> all draws and For=Against
(b) Sum(For)=Sum(Against)
(c) 0 pts -> WA lost all games
(d) J and WI have played everyone, so E has played >1game. Sum(Played) must be even (two teams per game), E = 3
(e) Sum(Points)=Sum(Played), Points€=5
(f) J and WI played everyone, so WA has played them but not played C or E
(g) C's ties are with J WI and E
(h) So WI won 1 and J won 2
8 games: J=C WI=C E=C E>WI W>J J<WA WI>WA J>WI

P	W	L	T	F	A	Points
4	2	1	1	7	1	5
2	0	2	0	2	6	0
4	1	2	1	5	10	3
3	0	0	3	4	4	3
3	2	0	1	5	2	5

Quote: Gialmere

...How were they able to make the correct calculation?...

Quote: Gialmere

It's easy Monday...

...

Four honest and hard-working computer engineers were having lunch together when the topic of salaries came up.

They decided they wanted to compute the average of all four of their annual incomes. None of the engineers, however, were willing to disclose how much money they actually made in a year.

How were they able to make the correct calculation?

Quote: charliepatrick

Show Spoiler

Just an idea that seems to work. Using two pieces of paper A writes down the same large value, known to exceed the value of the four salaries, and (can show them to B) and put one in the "original numbers" box and passes the other piece of paper to C. C subtracts their salary, writes the total on another piece of paper and hands it to D. D does the same and now puts their piece of paper in the "less two salaries" box. B, C and D similarly repeat the process. (It is even fairer if they all come up with their initial numbers at the same time and then randomise who gets which one. Another method is for people to come up with two initial random numbers each. Another method is only to use A and C to initiate the process, so divide by 1.)

Then calculate the difference between the total of "original numbers" and "less two salaries" and divide by 2.

Quote: ChesterDog

Show Spoiler

Four playing cards will be prepared, each with a different number on it. For example, the four numbers might be: 5,000; 10,000; 15,000; and 20,000.

The four cards are shuffled and dealt face down to the four engineers. Each adds his salary to the number on his card and announces the sum.

They add the four announced numbers, subtract 50,000 and divide by four to find their average salary.

Quote: Gialmere

....

Comment

As you can see, there are actually several ways to solve this puzzle that a clever person might think up. The "official" solve uses probably the simplest technique...

Engineer #1 secretly generates a large (say, seven figure) random number and adds his salary to the amount. He writes down the new number and gives it to engineer #2. She, in turn, secretly adds her salary to the number, writes down the new sum and passes it to engineer #3, and so on. When the final number returns to engineer #1, he subtracts from it the original random number and divides the difference by 4.

The engineers can now see how underpaid they are as a group.

...

Quote: Wizard

Just one solution and some thoughts

8,10,12.

I think any solution must have and odd number of even numbers (either 1 or 3).

It's easy to say that if the sides are x, y, and z, then 4*(xy+xz+yz) = xyz + 8*(x+y+z)

Quote: ThatDonGuy

I don't think these are all of them

Here are the ones I could find where all of the dimensions < 5000:
12 x 10 x 8
14 x 9 x 8
16 x 10 x 7
16 x 14 x 6
18 x 8 x 8
20 x 9 x 7
20 x 12 x 6
24 x 11 x 6
24 x 22 x 5
27 x 20 x 5
30 x 8 x 7
32 x 10 x 6
32 x 18 x 5
36 x 17 x 5
42 x 16 x 5
52 x 15 x 5
56 x 9 x 6
72 x 14 x 5
100 x 7 x 7
132 x 13 x 5
2581 x 2580 x 651
2836 x 2744 x 2218
2951 x 2176 x 1346
2988 x 2550 x 1700
3554 x 3216 x 2264
3570 x 3040 x 2781
4233 x 3222 x 3160
4479 x 2648 x 2182
4580 x 4501 x 1465
4700 x 3586 x 1536
4770 x 4279 x 2112
4831 x 3621 x 1972
4838 x 3606 x 2224
4858 x 4060 x 2840
4866 x 3840 x 3228
4962 x 4736 x 2750

Note that, if the dimensions are L x W x H, then LWH = 2 (L - 2) (W - 2) (H - 2)

I have a feeling there is no maximum solution, but I am still working on the proof (one way or the other)

Quote: Wizard

Here are my answers thus far

Quote: ThatDonGuy

EDIT: My earlier list had answers that were incorrect because of 32-bit integer overflows

Here are the ones I could find

12 x 10 x 8
14 x 9 x 8
16 x 10 x 7
16 x 14 x 6
18 x 8 x 8
20 x 9 x 7
20 x 12 x 6
24 x 11 x 6
24 x 22 x 5
27 x 20 x 5
30 x 8 x 7
32 x 10 x 6
32 x 18 x 5
36 x 17 x 5
42 x 16 x 5
52 x 15 x 5
56 x 9 x 6
72 x 14 x 5
100 x 7 x 7
132 x 13 x 5

Note that, if the dimensions are L x W x H, then LWH = 2 (L - 2) (W - 2) (H - 2)

Quote: Wizard

Here are my answers thus far. I don't think there are more and can't find any connection to them, other than a circular reference to the problem at hand.

Show Spoiler

5,13,132
5,14,72
5,15,52
5,16,42
5,17,36
5,18,32
5,20,27
6,9,56
6,10,32
6,11,24
6,12,20
6,14,16
7,7,100
7,8,30
7,9,20
7,10,16
8,8,18
8,9,14
8,10,12

Official Solve

Bad Joke

Quote: chevy

Follow up question:

For arbitrary rectangle with Height=H, Width=W......(oriented with H>=W), What range can the area ratio (Blue/Total) of the folded result have? And for what H,W is the Blue area 50%?

Quote: CrystalMath

Show Spoiler

I get a general equation for the ratio of H/W = sqrt((ratio+1)/(3ratio -1)) .
For a ratio of 0.5, H/W = sqrt(3)

If H=W, ratio = 1 (same as folding a square in half).
If H >> W, the ratio approaches 1/3.

Quote: Gialmere

Turkey = ?
Cornucopia = ?
Native American = ?
Mayflower = ?
Pilgrim = ?

Quote: chevy

Show Spoiler

***I "think" the red/blue diagram is not to scale. It makes blue appear as right angle. I "think" the line from A,D to the center of opposite side is the perpendicular.***

Call the center of the rectangle Y. Call the ends of the fold (X,Z). So side of blue triangle opposite A,D is XYZ with Y at midpoint and (A,D) to Y is perpendicular to XZ.

Area of Blue = .5 * XZ * AY = XY * AY
Area of rectangle = 2*Area of Blue + 2* Area of Red = 4
Area of Pentagon = Area of rectangle - Area of Blue = 4-Area of Blue

Ratio = Area of Blue / (4-Area of Blue)


From Trig using the rectangle diagram with XYZ added accordingly....
AD = sqrt(4^2+1^2)=sqrt(17)
AY = AD/2 = sqrt(17)/2

Call theta angle given by CAD, then
tan(theta) = DC/AC = 1/4

Theta is also angle for XAY, so
tan(theta) = XY/AY
XY = AY * tan(theta) = sqrt(17)/2 * (1/4) = sqrt(17)/8

Thus Area of Blue = XY * AY = [sqrt(17)/8] * sqrt(17)/2 = 17/16

And
ratio = Area of Blue / (4-Area of Blue)
= (17/16) / (4 - 17/16)
=17/47

Quote: ThatDonGuy

Answer

Just the numbers this time...

Turkey = 6
Cornucopia = 8
Native American = 7
Mayflower = 11
Pilgrim = 29

joke

Quote: Wizard

I agree!

Welcome to the WoV forum, by the way.

That was a tough problem and only one member answered it correctly (you), so please consider yourself invited to the prestigious "Beer Club." This means I owe you a beer should we ever meet.