Problem of the Week (POW)

November 25:

THE CONTEST is lonely! No entries have been received for this week's puzzler! So we'll give you another week. Think about it while you're digesting your pumpkin pie and send entries to jgalovich or under my door (PE 261). In case you have misplaced it, here it is again:

 

Alice and the Mad Hatter go home for Thanksgiving

 

``Two travelers spend from 3 o'clock till 9 in walking along a level road, up a hill, and home again: their pace on the level being 4 miles per hour, up hill 3, and downhill 6. Find distance walked; also (within half an hour) time of reaching top of hill."

 

(from Dodgson's A Tangled Tale (1885))

Contest entries accepted until noon on Monday, December 1. Send to jgalovich@csbsju.edu or put under my door (PE 261).

 

The figure shows a square, three circles, and some tangents. The two lower circles have the same size. Do all three have the same size?

November 10

The figure shows a square, three circles, and some tangents. The two lower circles have the same size. Do all three have the same size?

The conest will close at noon Monday November 17. As usual submit entries to jgalovich or put under my office door, PE 261.

Congratulations to Wenzhi Wang who has, once again, won this week’s contest. Thanks to all who entered!

 Upper division students – you are on notice!! Take up the challenge with the new contest (see below).

 

First, however, is Wenzhi’s solution:

 

We can mark these tickets from 1 to 200. And for the prizes we choose 100 different numbers from 200 tickets. Therefore, there are C¹⁰⁰₂₀₀ cases. When Bob has only one ticket, if the 100 numbers are chosen from other 199 numbers, he will not win the prize.

So the probability of winning is 1-C¹⁰⁰₁₉₉/C¹⁰⁰₂₀₀=1/2

Similarly, When Bob has two tickets, if the 100 numbers are chosen from other 198 numbers, he will not win the prize.

So the probability of winning is 1-C¹⁰⁰₁₉₈/C¹⁰⁰₂₀₀=299/398

So the ratio is (299/398)/(1/2)=299/199

October 31:

Alice and Bob are at a conference dinner and each of them holds a ticket for a prize drawing. There are 100 prizes and 200 tickets. The prizes are given by choosing tickets randomly; once chosen, a number is not eligible for a second prize.

 

Alice wants to leave so that she can watch a Blazers soccer game on CNN (OK, we could hope!) So she gives her ticket to Bob, saying: "Here, this will double your chances of winning something." Bob is not so sure. If there was only one prize, then holding two tickets would indeed double his probability of winning something. But if there were 200 prizes, then the probability of winning is 1, which is not improved by holding two tickets. When there are 100 prizes, what is the ratio of Bob's chances with two tickets compared with his chances with just one ticket?

 

This week's contest will close on Monday, 11/10 at noon. As always, entries to jgalovich@csbsju.edu or under my door at PE261.

October 24: You Could Win the Lottery!!!

All -- There were no winners this week, but Wenzhi was close. Here is his solution:

Since 000, 111, 222, ..., 999 we need only use 00, 11, 22, 33,..., 99 as the ticket for these. So these ten tickets are necessary. Moreover, 00, 11, 22,... can also win the number which has two of the same number, like 005, 838, 477 and so on. So there are only 9*8*7=720 numbers which has three different digits, like 125,758, 963 and so on. If we do not consider the order of the numbers: We just choose three different numbers from 0, 1, 2...9. So there are 9*8*7/3/2/1=120 different numbers. If we choose the first pair of ticket numbers like, 01, 23, 45, 67, 89, each number can represent 8 different three-digit numbers. So 5*8=40. Then we choose other 5 ticket numbers which don't have the same digits and are different from the first group, like 12, 34, 56, 78, 90. Because of the first pair of numbers, we can just get 5*6=30 three-digit numbers from the first group. Similarly, we choose the third group of five numbers, there will be 5*4=20 numbers different from the first and second group. Then there are 30 three-digit numbers left. 30/2=15. So we need at least 15 different numbers to win the remaining 30 three-digit numbers. so we need total 5+5+5+15=30 two-digit numbers to win 120 three-digit numbers without considering order. When we consider the order, we just reverse the order of the two-digit ticket. so we need 30*2=60 to win the 720 numbers. Plus 10 numbers like 00, 11, 22...99. We need at least total 70 tickets to win.

THE PROBLEM

Imagine a lottery where the winning number is abc, where letters are decimal digits and any or all of them can be 0. Thus there are 1000 possibilities for the winning number. But tickets for the lottery show only two digits, xy, and such a ticket wins if the winning number is xyA or Axy or xAy. In short, having two digits in the right order wins. There are 100 possible tickets to buy. How many tickets do you have to buy to guarantee a win?

As usual, entries can be submitted to jgalovich@csbsju.edu or leave them by my office (PE 261). Deadline is noon on Thursday, October 30, 2008.

October 17:A Quadratic Equation?xml:namespace prefix = o >

How many positive integers z are there so that x(x+z) = y^2 has no solutions in
positive integers x, y?

This week’s winner of the contest is Wenzhi Wang – congratulations Wenzhi! Wenzhi’s solution to last week’s contest:

(i) When z is an odd positive number (except z=1 ), suppose z=2k+1, (k is an positive integer). So z=(k+1+k)(k+1-k)=(k+1)^2-k^2. So we can find a positive integer x=k^2 so that x(x+z)=k^2 * (k+1)^2=(k^2+k)^2. So we can find y=k^2+k. Therefore, when z is 3, 5, 7, 9,11..., we can find at least a pair of positive integers x,y to satisfy x(x+z)=y^2

(ii) If z is an even number and has an odd factor (2k+1), (except 1) , then z can be written as z= (2k+1)*s, (s is an even number) (iii) From (i), we know that z=[(k+1)^2-k^2]*s, so we can find an positive integer x=k^2*s so that x*(x+z)=(k^2*s)*[(k+1)^2*s]=[k*(k+1)*s]^2. \ So we can find an integer y=k*(k+1)*s. Therefore, when z is 6,10, 12, 14, 18, 20, 22..., we can find at least a pair of positive integer x,y to satisfy x(x+z)=y^2

(iii) If z is an even number and has no odd factor (except 1), it can be written as z= 2^n, where n is a positive integer and n>2, then z=(3+1)*(3-1)*[2^(n-3)]=(9-1)*[2^(n-3)], so we can find an integer x=2^(n-3), so that x(x+z)=[2^(n-3)]*[9*2^(n-3)]=[3*2^(n-3)]^2. So we can choose y=3*2^(n-3). Therefore, when z is 8,16,32,64..., we can find at least a pair of positive integer x,y to satisfy x(x+z)=y^2 Through these three cases, we find z is not 1, 2, 4. We need to prove that when z=1,2,4 we cannot satisfy x(x+z)=y^2 Without loss of generality, assume x P>

Therefore, only when , we can find no solutions in positive integers x, y. So the number of possibilities is 3,namely z=1, z=2, z=4

October 10: Measure this Angle!!

No contest winners  last week, though Wenzhi Wang was very close!  

Suppose you wish to measure the angle made by two lines AB and AC using a ruleralone. Mark two points X and Y on AB, AC, respectively so that thedistances AX and AY are each m millimeters, m an integer. Measure thedistance XY in mm and use that number to approximate (in degrees) themeasure of angle BAC. What should m be so that this approximation isbest, in terms of minimizing the maximum error for angles of the form ndegrees, where n ranges from 1 to 90?

Solutions can be submitted to Jennifer Galovich in PE 261 or by email to jgalovich. Deadline for submission is noon on Thursday, October 16.

October 2: What is the Locus of the Focus? (No winners.) Solution:

Draw the line tangent to the graph at a typical point, say P=(x,x^2). Here's the idea: Rotate the graph about the point P until the tangent line is now parallel to the x-axis. Then lower the graph so that P is actually on the x-axis. Then figure out what your x coordinate is now. Since the slop of the tangent line is 2s at P, we know that the angel trough which the graph must rotated is arctan(2x). However, since the rotation is clockwise, this angle is -arctan(2x). How far do we have3 to lower the graph? That is just the y-coordinate of P, namely x^2. Finally where are we? The coordinate at which we have arrived is just the distance that the curve rolled, or the arclength of the curve from 0 to x. 

Now we want to track what happens to the point (0,1/4) under all this action. We write a function that describes what happens to any point (a,b) and then plug in a=0, b=1/4. In the end (Mathematica can help here) we end up with the following function f(x) = (1/4) Cosh(4x). The graph is a catenary, the shape of a curve that a hanging chain naturally takes. See picture:

Take the standard parabola y = x^2 with focus at (0, 1/4). If this parabola rolls without slipping along the x-axis, what is the locus of the focus.

Send your entries to jgalovich or put under my door (PE 261).  The contest ends on Thursday, October 9.

September 25: Get Your Goat!!

Congratulations to  Steven Rogers and Nick Matthees  , this week’s winners of the “Monty Hall” contest.  Thanks to Wenzhi Wang,  Jason Lutz and Francis Olsem who also submitted correct solutions.Here’s the solution provided by Steven Rogers:
 
After you make your first selection and the host opens a door, showing a goat, you would still stay with the first one you chose even though you still only have a 25% chance, but you also know that the host is going to show you one more door that has a goat. Now, after you stay, and the host opens another door, showing another goat, you would then make a switch, taking a 75% chance instead of the now 25% chance you had left.

Background -- The infamous Monty Hall problem: There are three doors, with a car behind one door and a goat behind the others. I pick, Monty opens a door showing me a goat and ask me if I want to switch. I should switch, as that changes my probability of getting the car from 1 in 3 to 2 in 3.

Suppose there are four doors. After my first choice Monty opens a door, shows me a goat, and offers me a chance to switch with one of the two unopened doors. After I make my decision, Monty opens a door, shows me a goat, and offers me a chance to switch to the remaining closed door. What should my two decisions be?

Note: I know in advance that Monty will follow this protocol, showing me two goats, regardless of choices I make, and making his choice randomly.

September 19: A Number Trick

Congratulations to this week’s winners, Wenzhi Wang and Jason Lutz and thanks also to Brendon Murn, Vince Wheeler, Fru Nde, Vanessa Kvam and others for their solutions. Here is Wenzhi’s solution: Suppose the original number is 10000a+1000b+100c+10d+e, which a, b, c, d, e are all chosen from 0,1,2,3,4,5,6,7,8,9. So 10000a+1000b+100c+10d+e-(10000e+1000d+100c+10b+a) =9999a+990b-990d-9999e =9999(a-e)+990(b-d) =99〔101(a-e)+10(b-d)〕 =9*11*〔101(a-e)+10(b-d) We know that the four given numbers are 1, 1, 0, 8. I suppose the circled number is x. The five-digit number has the factor 9. If it can be divided exactly by 9, the sum of the five-digit numbers must be divided exactly by 9. So the remaining number must be 8.

Alice: Pick a 5-digit number.Bob: Got it.Alice: Now reverse it and subtract it from the original.Bob: Done.Alice: Now pick one of the five digits of the result and circle it (but don't choose a zero). Tell me the other four digits, in some random order.Bob: Sure; they are 1, 1, 0, and 8.Alice: Then the circled digit is....

Alice does indeed correctly give the circled digit. How did she do it?

September 12: Jokers Wild

 

 

Considera 54-card deck consisting of the usual 52 cards and two jokers, whichare wild, meaning that they can represent any of the 52 standard cards.For a standard deal of five cards from the 54, which poker hand is morelikely: four of a kind or a full house (three of a kind and a pair)?

 

Note: "Four of a kind" refers to atrue four of a kind, and excludes five of a kind.  Also, for this puzzle, each hand is assigned to its optimal possibility according to the rules of poker. So  AAA2 Joker mustbe taken to be four of a kind (not a full house). And 3 3 3 3 Joker would count as 5 of a kind, not 4 of a kind.

 

 

September 3: It's a Piece of Cake!!

The Piece of Cake problem was not solved correctly by anyone.  It IS pretty sneaky. (Moral: DRAW THE PICTURE!!!)

MANY people, including several professionalmathematicians, thought the answer was 720, which is also the answergiven by all student entries. However the correct answer is 4. See thefigure below:

 

 

A round cake has icing on the top but not the bottom. Cut out a piece in the usual shape (sector of a circle with vertex at the center), pull it out, turn it upside down, and replace it in the cake to restore roundness. Do the same with the next piece; i.e., take a second piece with the same vertex angle, but rotated counterclockwise from the first one so that one boundary edge coincides with a boundary edge of the first piece. Remove it, flip it, and replace it. Keep doing this in a counterclockwise direction. The diagram below shows what happens after two moves in the case that the central angle is a right angle.   For some central angles what happens is clear. For a 90 degree angle it takes eight moves for all the icing to return to the top. If the angle is 180 degrees, it takes four moves to return to the initial state. Suppose the central angle is 181 degrees. When does all the icing first return to the top?

 

August 27: Get Naked!!!

Contest Winners for August 27: 1st place: Zach Dimond 2nd place: Ha Pham and Nga Tran

This is an unusual Sudoku puzzle called Naked Sudoku, because, as you can see, the puzzle is not wearing any entries to start with.  Your job is to fill in the puzzle, using the usual Sudoku rules: digits from 1 to 9, no two the same in each row, column, or 3x3 subsquare). The small > and < signs indicate which square has the larger number; thus the entry in the upper left is smaller than its right neighbor and larger than its lower neighbor. As an aid, any square that is smaller than all its neighbors is colored blue; any square that is larger than all its neighbors is colored pink.

 

Paper copies can also be found outside Jennifer&rsquos office door (PE 261). Put your name and the date and time of submission on your solution and put it under her door.  Winners will be announced. Prizes may be awarded.