Random questions from my math class

Note: I posted these somewhere else so if you already saw them and the answer, please don’t ruin it by just posting the answer here. Also, the list here isn’t a complete list of the other one as some I decided not to include due to certain factors. If there’s a problem due to color schemes between the 2 forums, or anything else, just make note of it and I’ll fix it. Also some questions may have an inside joke or whatever due to the other forum so if you don’t’ get it, don’t worry. (Other forum is a guild site)

Also as I started the original thread back in January, it may be confusing when I mention something time related like “I’ll post the answer in February”. Just ignore it or ask if something isn’t clarified enough.

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

Apparently we get a Puzzle question every class meeting, tuesday/thursday, so i'll post them here and see what people respond to them as. Some things may not be a question from the class but i'll formulate it into one. Was quite interesting. It's a combinatorics class which is basically permutations, combinations, counting (General word: Enumeration) and some other things like Latin squares (Lol sudoku).

I'll post my solution to the problem when I post a new puzzle or if someone gets it right/close, whichever comes first.

 ”Question #1

1. A tennis tournament has 153 people participating. Matches are played elimination style where each match consists of 1 person against another. The winner goes onto the next match.

For example:

Round 1: 76 matches are played. 1 person does not play and advances automatically. 77 people left
Round 2: 38 matches are played. 1 person does not play and advances automatically. 39 people left

etc...

While it is easy to count this, provide a solution/approach to this problem that simplifies it and explain your reasoning. There might be a better reason than that one we received in class but I'll see if it is or not.

Main question: How many matches are played?

 ”Question #2

2. I hand Dean 10 playing cards, ace through 10. I provide him a deal that if he can arrange the 10 cards in all possible arrangements in a month, I shall give him nudes of Xtina. He can write them down, type them, anything as long as they are not automatically generated. If he cannot complete the task, he must post a picture of himself on these forums.

Should Dean accept my offer? Explain why or why not.

 ”Question #3

3. You have 2 jars.

- Jar #1 has 200 green marbles
- Jar #2 has 200 red marbles

- You transfer 30 green marbles from Jar #1 to Jar #2
- You thoroughly shake Jar #2
- You then transfer 30 marbles from Jar #2 to Jar #1

Are there now more red marbles in Jar #1 or Green marbles in Jar #2? To ask it another way, which is greater, the amount of red marbles in Jar #1 or green marbles in Jar #2

 ”Question #4”

4. You have a 2^n * 2^n checkerboard like the one below

______________________
l__l__ll__ll__ll__ll__ll__ll__l
l__ll__ll__ll__ll__ll__ll__ll__l
l__ll__ll__ll__ll__ll__ll__ll__l
l__ll__ll__ll__ll__ll__ll__ll__l
l__l__ll__ll__ll__ll__ll__ll__l
l__ll__ll__ll__ll__ll__ll__ll__l
l__ll__ll__ll__ll__ll__ll__ll__l
l__ll__ll__ll__ll__ll__ll__ll__l

The above is an 8 * 8 and has 64 pieces. Whenever one square is removed, the rest of the board can be tiled with the following tile

__
l__l__
l__ll__l

Where that piece can be rotated much like in Tetris. This is easily seeable with the 2x2 case. Prove that the bolded statement above is true. I'll post a hint to this problem over the weekend if no one has solved it.

 ”Question #5”

5. It is raining outside and you have 'N' people enter a restaurant, each carrying an umbrella. They drop it off at the front. When they leave, they collect an umbrella back, which may or may not be their own. What is the probability that no one got their own umbrella back?

 ”Question #6”

6. Dean is cheating on Fawn and is dating 8 girls at once (Names chosen at random)

- Xtina
- Anna
- Kimmy
- Ling
- Debbs
- Pog
- Marc
- Ronnie

He has a phone number for each of them and wrote them on separate pieces of paper. Unfortunately being the noob he is, he did not write the name of each girl by the phone number. Here are the 8 numbers

1-800-Horny4u
1-800-iWant69
1-800-SexyFun
1-800-HawtAzn
1-800-LazyHoe
1-800-MorePlz
1-800-HumpYou
1-800-iRwhore

He randomly assigns the phone numbers to the girls and concludes that:

A. The chance that all the numbers are for the correct girl is 1/8!

B. The chance that none of the numbers are for the correct girl is close to (exact answer taken out but 37%~)

While doing this math, he forgot to shut the door and Fawn walks in on him and finds out what he has been doing. Instead of being enraged, she offers Dean a deal. She knows which number belongs to each girl as she is friends with all of them. She also informs dean that only 5 of them are girls and dean is very shocked of this news. She proceeds to ask ask him a question. If he answers it correctly, Fawn will call the 5 girls over for an orgy. If he is incorrect, she will give dean the numbers to the 3 guys and he will have a very unpleasant night.

Question: What is the chance that exactly 7 of the numbers assigned belong to the correct person?

 ”Question #7”

7. A school encyclopedia has 10 volumes each with 100 pages numbered consecutively from 1 to 100. When they are on the library shelf (in the usual way) how many of the numbered pages are between page 1 and page 1000

Note: “Between” means that, the # of pages between page 4 and page 6 is only 1 page.

Note 2: For clarification


Volume 1: Pages 1~100
Volume 2: Pages 101~200
.
.
.
Volume 10: Pages 901~1000

 ”Question #8”

8. Gonna parphrase this one since it's wellknown. Please don't just google and paste the answer -.-. It's an adaptation of it though

You have a prison with N cells, each one closed. A guard walks past certain cells and either opens or closes it, depending on the current situation

If it's opened, the guard will close it
If it's closed, the guard will open it

Guard #1 walks past cell #1, cell #2, ... , cell #n, which are the multiples of 1
Guard #2 walks past cell #2, cell #4, ... , cell #2n which are the multiples of 2

This pattern continues with each guard #x opening/closing the multiples of x

i.e. Guard #7 opens/closes #7, #14, #21, etc...

Which cells are left opened. Since it's easy to spot the pattern, explain why.

 ”Question #9”

9. There are 3 index cards in Dean's hat

- One has nudes of (Insert girl's name here) on both sides
- One has a tease picture of (Insert girl's name here) on both sides
- One has a nude of (Insert girl's name here) on one side and a tease picture of her on the other side

Dean selects one card at random and places it on the ground so only 1 side is visible. The visible side is a nude of (Insert girl's name here). What is the chance that the opposite side of the card is also a nude picture?

 ”Question #10”

10. It’s the monty hall problem of doors and switching and I will not be discussing this one here f3]. There’s already a separate thread for it

 ”Question #11”

#11: An eccentric billionaire has invited you to play the following game:

- He selects a positive real number "at random"
- He writes it in decimal notation (i.e. 387.420489) on a card, and on a second card he writes twice the value of the card (i.e. 774.826978). The numbers are interpreted as dollars and cents, where the leftover and just be disgarded or rounded

Both cards are face down. You are allowed to look at only one card and you see a certain value (Denoted as "Y" for this problem).

You can get ask to receive either this value or the value on the unseen card. You will get the amount of money on the card you choose. What should you do?

Paradox within the problem:

When you see the card and get the value "Y", if you switch you can either get "Y/2" or "2Y". On average, that equates to

(Y/2 + 2Y)/2 = (5Y)/(4)

Meaning switching on average yields a 25% increase of the card you have chosen. However, that works for whatever card you got, whether it be the original value or the x2 value card. Obvioulsy there is something wrong here. What is wrong?

 ”Question #12”

12. I liked this one.

You have 10 jars of coins, each with 50 coins.

A "good" coin weighs 10 grams and a "defective" coin weighs 9 grams. It is suspected that one jar may contain all defective coins.

You have a scale (Where there is only one spot to weigh coins similar to a post office/bathroom scale, not a balance scale) that will way, in increments of 1g, from 0 to 600 grams.

What strategy will determine, with the fewest weighings, if any jar has defective coins and if so, which jar?

Note: You can assume the jars are weightless or whatever

Note 2: The scale will only give you a weight if you press a button. You cannot simply get a weight, and add more weight to it and still count that as “1 weighing”

Note 3/Clarification: Say all 10 jars contain good coins

Jar #1 has 50 good coins, weighs 500 grams as 50 * 10 = 500
Jar #2 = same exact thing
.
.
Jar #10 = same exact thing

Say Jar #2 contains defective coins

- ONLY Jar #2 will contain defective coins
- All coins in Jar #2 will be defective

If you weighed all jars except #2, they will be 500g each. If you weigh all coins in Jar #2, it will weigh 450g due to 9g * 50 coins

 ”Question #13”

13. There are 15 billard balls, numbered 1~15, on a pool table

Rules:

- You may knock off any one of these balls to start, but each subsequent ball must have a number adjacent to one already off the table

Example:

If ball #5 was first ball off, then the next one must be either #4 or #6. If the first two off were #5 and #4, the next one must be #3 or #6

Ball #1 is adjacent only to ball #2
Ball #15 is adjacent only to ball #14

This means ball #1 and #15 are not "connected" like #3 and #4 are, #6 and #7 are, etc...

How many possible sequences are there for all 15 billard balls to go off the table?

Note: There's a "hard" way to do this and a "trick". I doubt anyone will get the trick so I'll explain it over the weekend or whenever I have time after someone gets the answer.

 ”Question #14”

14. The "Gregorian Calendar" is the current calendar most people are used to. It follows these "rules"

1. An "ordinary year" has 365 days
2. A "leap year" has 366 days

3. A year is ordinary unless the year number is a multiple of 4**, in which case it is a leap year unless the year number is a multiple of 100, in which case it is not a leap year, unless the year number is a multiple of 400, in which case it is a leap year.

That's a fun fact btw. I never knew it myself

Examples:

1663 = ordinary year because it is not a multiple of 4
1976 = leap year because it is a multiple of 4
2000 = leap year because it is a multiple of 400
1900, 2100 = ordinary year because it is a multiple of 100

So if anyone actually makes it to the year 2100, it'll be a non-leap year :x

**If you stop the "rules" where this symbol is, those are the rules for the Julian calendar (Fun fact, not needed for problem)

Question: Prove that the probability that Christmas (December 25th) falls on a Wednesday is not equal to 1/7

Hint/Info from professor: I could have chosen any day of the week and it would have not made a difference to the problem. Likewise, I could have chosen January 1st (Somewhat pointless as new years and christmas land on the same day but you get the point that any day could have been selected) and it still would not change anything.

 ”Question #15”

15. OCT 31 = DEC 25

Find an explanation.

 ”Question #16”

16. Polar Explorer Xtina starts out at 8am from her base camp, goes 40 miles due South, has lunch, goes 40 miles due East, has coffee, goes 40 miles due North, and finds she is back at her base camp.

A. Where is her base camp?

If you solved A:

B.

 Spoiler

Same scenario as Part A except she has never traveled north of the Arctic Circle

 ”Question #17”

17. Explorer Dean planned a trip to Antartica to see Fawnie that would to last from May 1st --> Nov 30th. He would travel by boat.

His guildmates were being superstitous and wanted to avoid any Friday the 13ths. In any 1 month, the probability of no Friday the 13ths is roughly 6/7

So in a span of 7 covered months (Which is May 1st --> Nov 30th), the chance of getting of getting no Friday the 13ths is roughly (6/7)^7 or about 34%

Thusly, in any one year, the chance of getting a Friday the 13th is 66%

For example, in 10 years, the chance of having a "favorable year" (A year with no Friday the 13ths in a 7 month span) should have the probability of

1 - 0.66^10, which is roughly 0.999+

Dean has been waiting 45 years and still has not had a favorable year. Fawnie has already given up waiting and explained to Dean why he has not had a favorable year. What is her answer?

 ”Question #18”

18. Difficulty is as follows

Part A: Easy
Part B: Medium, has 3 inequivalent solutions
Part C: Very difficult, has 1 unique solution

Inequivalent means solutions that are rotations, symmetry, etc of another are the same solution

A) You are given the following picture

[ . ][ . ][ .
[ . ][ . ][ .
[ . ][ . ][ .

Draw a continuous line of 4 line segments that goes through all 9 points

B) You are given the following picture

[ . ][ . ][ . ][ .
[ . ][ . ][ . ][ .
[ . ][ . ][ . ][ .
[ . ][ . ][ . ][ .

Draw a continuous line of 6 line segments that goes through all 16 points and finishes where it started.

C) You are given the following picture

[ . ][ . ][ . ][ .
[ . ][ . ][ . ][ .
[ . ][ . ][ . ][ .

Draw a continuous line of 5 line segments that goes through all 12 points and finishes where it started.

 ”Question #19”

19. Forgive me for the diagrams as I tried my best. Each picture has 2 goals

Goal #1: Partition the figure into 4 congruent pieces that are similar to the original

Partition - Imagine creating invisible or dotted lines that "cut" the diagram into sections

Congruent - Each section is identical to a previous section, except it might be rotated

Similar - Each section is proportional to the original figure

Goal #2: Cut each figure into 9 congruent rectangles (This does not apply to the example nor to figure A)

The pictures may have built-in lines to illustrate dimensions of the figure. You do not have to confine yourself to those lines as they are just a visual aid to understand what the picture looks like.

Example: The original image on the left and the answer on the right



Figure A: A rectange divided into 4ths and only 3 remaining



Figure B: A rectange divided into 6ths and 4 remaining



Figure C: A rectange divided into 6ths and 5 remaining



Figure D: A specific looking trapezoid (Square and triangle)



Figure E: The top half of a hexagon



Figure F: The top half of a hexagon cut in half (Diagonal and base are of length "a", height is length "h" with top being "half of h"



Figure G: A triangle + a paralleogram (See if you can find a name for it ;-

 ”Question #20”

20.



Both polygons are squares with one of them having a side of length 4 feet and the other having a side of length 7 feet. The bigger square intersects the smaller square at 16 inches [1/3 of 4 feet]

Find the area of the overlapped section (I was suppoed to shade it but forgot so w/e) and use as little reasoning as possible.

P.S. After looking at the preview, there are a few slight errors such as the quotes in the spoiler tag but nothing big enough so a problem cannot be finished. Please put your answers in spoilers so people won't see them if they don't want to see them

 1

1. Number of people leaving = number of matches played. Therefore, total number of people - total matches = number of people left. So if there are x people in a tournament, there are x-1 matches played. 152 matches are played.

Will be editing answers in.

 2

The number of total permutations that cards can be arranged in is 10! (= 10 * 9 * 8 * .... * 1), which is roughly 3.6million combinations. That would require him to do rougly 360,000 a day, or 6000 an hour (without sleep), or 100 a minute. This is impossible, therefore he should not take the deal.

 3

When 30 are transfered to jar 2, there is a 100% chance that they will all be green. When 30 are transfered back, there is no such guarantee that they will all be red. Therefore, there will be more green in jar 2 (30) than red in jar 1 (<30)

 5

for n umbrellas, each person has a (n-1)/n chance of not getting his/her umbrella back (1/n for getting their umbrella back). Therefore, for n people, the probability is ((n-1)/n)^n that no one will get their umbrella back.

 7

800. The way books are in shelves normally, page 1 will be at the right-hand side and page 100 will be on the left hand side (from your perspective). The reverse for pages 901 and 1000. Therefore, only the pages from books 2-9 will be between page 1 and page 1000

 8

Doors with numbers that can be square rooted evenly will be left open (ie 1, 4, 9, 16, etc.). Only these numbers have an odd number of factors, hence the state of the doors is changed an odd number of times, leading its final state to be the opposite of what it started out as.

 9

I'm going to say 50%, but I'm generally bad with these types of questions :x

 14

Lazy proof: In two consecutive normal years, every date will be moved over by one (ie if Christmas was on Wednesday, next year it'll be on Thursday). In a leap year, dates skip a day, so if Christmas was on Wednesday last year, it'll be Friday this year. Since there's a high chance of there being a leap year in a span of 7 years, there is a chance that it will not fall on a Wednesday, and also chance that it'll fall on a Wednesday twice. Too lazy to find the real probability. :p

 16

A. North Pole
B. South Pole? Is it even possible to go South from the south pole?
Btw, I ignored the fact that magnetic north and south poles don't necessarily line with with the geographical ones.

 ”Question #20”

20.



Both polygons are squares with one of them having a side of length 4 feet and the other having a side of length 7 feet. The bigger square intersects the smaller square at 16 inches [1/3 of 4 feet]

Find the area of the overlapped section (I was suppoed to shade it but forgot so w/e) and use as little reasoning as possible.

Just wanted to point out that this question is impossible to solve, because you can rotate the squares from 0 to 180 degrees and that changes it's area.

 #4

By induction...
Base case is true (2x2) as you showed.

Assuming for a 2^k x 2^k board, you can fill it leaving out one corner of the board:
Create a 2x2 array of those boards, where 3 have the corner facing the centre and 1 has it facing out:

Code:

BB|B0
B0|BB
-----
B0|0B
BB|BB

Place another L into the whole in the middle and you have a 2^k+1 x 2^k+1 with 1 corner piece missing.
Induction complete.

 #6

If 7 numbers are right, the 8th can't be wrong. So 0.

 #11

I don't see what the paradox.
There are 2 equally likely options for a given revealed card Y (assuming all equal sized ranges of real numbers are equally likely - that is, uniform)
2/Y, Y
Y, 2*Y
It could be considered payment for risk - if you stay, you have a 100% chance of keeping your value. If you switch you could lose it, or you could profit.

 #12

I'm assuming you can take coins out of the jars to weigh them.
If not it's kinda trivial - weigh each jar, find whether they are 500g or not, and done with it. If you weigh more than one at once, then you exceed the 600g limit.

Anyway, assuming you can take them out...
Weigh coins from 7 jars.

• None fake: Weigh 2 of the other jars
  • None fake: Weigh the last jar
    • Not fake: none are fake
      
    • fake: You know it's this jar
  • One of the 2 is fake: Weigh one.
    • Not fake: It's the other one.
      
    • Fake: You know it's this jar
• Fake in here: Weigh 4 of the 7.
  • None fake: Weigh 2 of the remaining jars
    • None fake: It's the other one.
      
    • One is fake: Weigh one of the jars.
      • If it's fake, you know it.
        
      • If it's not fake, it's the other one.
  • One is fake: Weigh 2 of these jars.
    • If one is fake, weigh one of them.
      • If this one is fake, done.
        
      • If this one is not fake, it's the other one.
    • If none is fake, weigh one of the other 2.
      • If it's fake, done.
        
      • If it's not fake, it's the other one you didn't weigh.

After 3 weighings you only have 8 possible outcomes (2^3) so you can't always determine which one is fake.

 #13

After much derivation

Code:

1 - 1 way

= 1

2: 3: 4: 5: (...) 14: 15 - 1 = 14 ways
            (...)     1 - 1
         1 - 1
      1 - 1
   1 - 1

= sum(1 to 14) 1


3: 4: 5: 6: ... = 13+12+11+...+1 = (13+1)*13/2 = 91 ways
         2: 6: (...) 15 = 11
            1 - 1
      2: 5: 6: (...) 15 = 12
            1 - 1
         1 - 1
   2: 4: 5: 6: (...) 15 = 13
            1 - 1
         1 - 1
      1 - 1

= sum(i = 1 to 13) sum (j = 1 to i) 1

4: 5: 6: 7:
         3: 7:
            2:
      3: (...) = (11+1)*11/2 = 66 ways
   3: 5: 6: 7: = (12+1)*12/2 = 78 ways
            2: 7:

         2: 6: 7:

      2: 5: 6: 7: ... = 12 ways
            1 - 1
         1 - 1

= sum(i = 1 to 12) sum (j = 1 to i) sum (k = 1 to j) 1

Basically
sum (i = 1 to K) sum ( sum( sum ( ... (15-K times) 1)))) = K'th ball
This works out to be
1 14 91 364 1001 2002 3003 3432 3003 2002 1001 364 91 14 1
= 16384

This number is familiar to me (2^14) and makes me wonder what the trick is to rearrange them so they're easier to add up.

 #20

4 square feet.

Just wanted to point out that this question is impossible to solve, because you can rotate the squares from 0 to 180 degrees and that changes it's area.

Are you scared of making equations that will tell you the area of the cross sections based on degree input? In the real world, people won't put up with slackers. (don't tell me those problems don't exist)

I do agree that question needs more clarity.

-readdresses some of the already tackled problems-

 (9)

Conditional probability: P(A given B) = P(A and B)/P(B)
Since there are 6 sides, 3 of which are nude, and the side that is visible is chosen at random, there is a 1/2 chance of the visible being nude [P(B) = 1/2].
The probability that both sides are nudes (and, following from that, the visible is also nude) is 1/3.
Therefore, P(other_nude given visible_nude) = (1/3)/(1/2) = 2/3.

 (11)

The problem is that the distribution of Y is assumed to be a uniformly random positive number. This also means that Y's average value is infinite. Infinity*5/4 is still infinite, so that conclusion would indeed hold. But, all this is illogical anyway because no one has an infinite amount of money to give.

 (13)

Say you start off with the billiard ball numbered n. There are n-1 balls less than n, and 15-n balls greater than n.
Due to the restrictions on the problem, once you've selected your first billiard ball, there are only two legal "moves" at any point: knock off the ball numbered closest to but less than n (let's abbreviate this action L), or knock off the ball numbered closest to but greater than n (abbreviated R).

Therefore the number of moves starting with n, is simply the number of ways to permute n-1 instances of L and 15-n instances of R, or 14 choose n-1.

Summing up the 15 possibilities for the starting value of n, this is sum(i = 0 to 14) 14 choose i, which equals 2^14 = 16384.

 (14)

Every 400 years, there are 100-4+1 = 97 leap years, and 303 normal years. This is 97*366+303*365 = 146097 days. Since this divides into 7 evenly (146097/7 = 20871), the probability that any given day in this 400-year span is 1/7. This means that this 400-year span can (excuse the non-technical language) be copypasta'd to the rest of history without biasing the distribution of weekdays.
However, in 400 years, there are 400 Christmases. 400 cannot be evenly divided into 7, therefore the probability that any given Christmas in the 400-year span (and therefore the rest of history) falls on a Wednesday cannot be 1/7.

 (16B)

Her base camp is 40 miles away from the south pole in any direction. Going "east" at the south pole amounts to spinning in circles.

 15

31 in octal (base 8) is 25 in decimal (base 10). 3*8+1 = 24+1 = 25.

 17

Not having a Friday the 13th in a month is not an independent event. I would assume from the given information that they occur approximately once every 7 months.

Would calculate, but meh.

 19

A B
C D
E F
G G

^way2smart lol

#20 stereo got right but you're basically looking for the area of the overlapped secction. Had no reasoning though f3

I think everything was right except for #2 with the marbles [unless someone went and re-tried it]. I just know the first attempt at it was incorrect.

Oh, and the jar one was too [incorrect]. You can take coins out of the jar.

#20 stereo got right but you're basically looking for the area of the overlapped secction. Had no reasoning though f3

In order to get Stereo's answer, you would have to assume that C is the center of the 4 inch square. <~ Reasoning

Otherwise, since there's no definite intersection point, you can rotate the square in any direction, and slide it back and forth and the answer would be 0 < area < 16. (If you've taken geometry, you should know about this.)

Shading the overlapping has nothing to do with clarity because it's clear from your diagram where they overlap.

Musta forgot to mention, but yes C is the center of the square.

You can take coins out of the jar.

Yeah, I wasn't thinking about that one.
10 from jar 1
9 from jar 2
...
1 from jar 10

0 gram (550 grams) = all good
-1 gram = jar 10 is fake
-2 grams = jar 9 is fake
...
-10 grams = jar 1 is fake

@ #3 (you said 2) they're identical.
# of red in jar 1 = # of green in jar 2
It has to be like this or there wouldn't be conservation of marbles.