# Fall 2008

Mathematical puzzles for your intellectual amusement.

December 4 Problem. It is our wish that HOPE and LOVE will give us PEACE. At least, if each letter represents a digit, we can make HOPE + LOVE = PEACE. And could there be more than one answer?

Solution by Nick Bouwman. The possibilities are obtained by looking at constraints, such as E must equal 0 to get E+E=E. P=1 because its in the beginning and even 9+9 is not going to create a number in the 20′s. Since P+V=C and P=1, C will be one more than V. H+L must equal 10 or 9 to get the PE which is 10. If H+L=9 then 0 must be at least 5 in order to have a one carry over.  In all cases, H and L are switchable. So below are all the possibilities, but times two since in each case H & L can be switched.  HOPE plus LOVE does indeed bring PEACE in a total of 32 different ways.

 E=0 P=1 O=2 A=4 V=5 C=6 H=7 L=3 O=2 A=4 V=8 C=9 H=7 L=3 O=3 A=6 V=4 C=5 H=8 L=2 O=4 A=8 V=5 C=6 H=7 L=3 O=6 A=2 V=7 C=8 H=4 L=5 V=8 C=9 H=4 L=5 O=7 A=4 V=8 C=9 H=3 L=6 O=8 A=6 V=3 C=4 H=2 L=7 V=4 C=5 V=2 C=3 H=4 L=5 O=9 A=8 V=3 C=4 H=2 L=7 V=4 C=5 V=5 C=6 V=4 C=5 H=3 L=6 V=2 C=3 H=4 L=5 V=6 C=7 H=4 L=5

November 20 Problem. At the Thanksgiving Buffet, there are three entrees (turkey, ham and lentil loaf), 11 sides (turkey stuffing, walnut stuffing, mashed potatoes, sweet potatoes, green beans, carrots, corn, relish, green salad, cranberry sauce and fruit cup), and four desserts (pumpkin pie, apple tort, pudding and chocolate cake). Customers choose five items, up to two being entrees and up to one being a dessert. How many different dinners can a customer obtain?

Solution by Seth Unruh. If we assume they person has to choose 5 items, then there are so many options. They can choose 2 entrees with 2 sides and one dessert. or two entrees with 3 sides and no desserts or one entree, 3 sides and one dessert, or one entree, four sides and no desserts, or no entrees 4 sides and one dessert or no entrees 5 sides and no desserts. All this to say you can use this information to figure out there are 5907 possible combinations.  Here is the calculation:

November 13 Problem. A bubble with radius 5 cm adheres to a bubble with radius 10 cm. Is the surface where they adhere (1) flat, (2) pushed in towards the bigger bubble, or (3) pushed in toward the smaller bubble? Why?

Correct Physical Solution by Stan Kaufman (’62) . At the midplane of a bubble of radius R, the surface tension produces a force 2*pi*R*s holding the hemispheres together, which is balanced by the pressure in the bubble.  The area of this midplane is pi*R*R, so the pressure (force/area) is 2*s/R.  That means the pressure in a small bubble is higher than that in a big bubble.  So the interface will be pushed in towards the bigger bubble [in order to equalize the pressure].

When you blow up a balloon, it is always much harder to start it than to continue once it’s started.  That’s what occurred to me first, but then there’s the matter of this nonlinear material that has to be stretched, so I didn’t trust the analogy.

Incorrect but Most Creative Solution by the Accounting Office.  The answer is #1 flat. Why? Because the elevated hot air concentration in our office, when applied to the Miracle Bubbles, along with the organic anomaly contained in the aura of greatness associated with the superior genetic composition of deoxyribonucleic acid transmitted genetically to the personnel of the office at birth, the bubbles cling together equally in desperation.

Comment by David Housman. Empirical proof can be seen at http://en.wikipedia.org/wiki/Image:Ggb_in_soap_bubble_1.jpg.  A discussion of the mathematical proof can be found at http://www.nsf.gov/discoveries/disc_summ.jsp?cntn_id=100596&org=NSF.

November 6 Problem. The Electoral College, not the popular vote, determines the next President. This year the distinction does not matter because Barak Obama obtained both a majority of the popular and the electoral votes. However, in 2000 Al Gore was the popular vote winner (by about a half-million votes) while George Bush was the electoral vote winner (by five votes). What is the smallest fraction of the popular vote a candidate could receive and still win the electoral vote?

Solution by Seth Unruh. In order to win the electoral votes and become president you need 270 electoral votes out of 538. So if each of the 270 were won by only one vote, that would mean basically candidate A who won got basically 50% of those votes, and the other 268 electoral votes were won by candidate B by 100% . So assuming the population is distributed equally over the electoral votes then they winning candidate would have received 135/538 of the popular vote, which is around 25%. So the Electoral vote winner could only get 25% of the popular vote and still win.

Comment by David Housman. Of course, Seth’s argument can be taken to an extreme. If only one voter votes in states that have a majority of the electoral votes, and those very few voters all vote for our candidate, while all registered voters in the remaining states vote for the other candidate, our candidate will win the Presidency with almost 0% of the popular vote.

October 30 Problem.
455 + 766 = 1332
455 + 766 = 1221
455 + 766 = 1110
are all true statements! How can that be?

Solution by Glenn Gilbert.
Base 9, Base 10, and Base 11.

October 23 Problem.
Larry, Curly, and Mo were the three candidates for Stooges Village Mayor. Each of the 20 citizens was to list the three candidates in order of preference on her or his ballot. When the ballots were opened, it was determined that 11 voters preferred Larry over Curly, 12 voters preferred Curly over Mo, and 14 voters preferred Mo over Larry! How many first place votes did Larry receive?

Solution by Staci Locke.
Larry received 6 first place votes and Curly was elected mayor by a landslide.

How I did it, from this table:

I gave Larry 11-1′s and 9-2′s while giving Curly the opposite.

I then gave Mo 12-3′s from the bottom up. Since I needed the rest of the Mo votes to be better than Curly, I gave Curly the 3 and Mo the 2 for the top eight voters.  This preserved my Larry/Curly ratio and completed the Curly/Mo ratio.

Then for the Mo/Larry relationship I again started form the bottom up and switched Larry’s-2 and Mo’s-3 which gave me 9 votes for Mo over Curly. Still needing 5 votes I went to the top of the table and switched Larry’s-1 and Mo’s-2 for the other five votes while not changing the Larry/Curly or Curly/Mo ratios.

Voter Larry Curly Mo Voter Larry Curly Mo Voter Larry Curly A 1 2 A 1 3 2 A 2 3 1 B 1 2 B 1 3 2 B 2 3 1 C 1 2 C 1 3 2 C 2 3 1 D 1 2 D 1 3 2 D 2 3 1 E 1 2 E 1 3 2 E 2 3 1 F 1 2 F 1 3 2 F 1 3 2 G 1 2 G 1 3 2 G 1 3 2 H 1 2 H 1 3 2 H 1 3 2 I 1 2 I 1 2 3 I 1 2 3 J 1 2 J 1 2 3 J 1 2 3 K 1 2 K 1 2 3 K 1 2 3 L 2 1 L 2 1 3 L 3 1 2 M 2 1 M 2 1 3 M 3 1 2 N 2 1 N 2 1 3 N 3 1 2 O 2 1 O 2 1 3 O 3 1 2 P 2 1 P 2 1 3 P 3 1 2 Q 2 1 Q 2 1 3 Q 3 1 2 R 2 1 R 2 1 3 R 3 1 2 S 2 1 S 2 1 3 S 3 1 2 T 2 1 T 2 1 3 T 3 1 2

David Housman Comments. Staci’s solution shows that Larry could have received 6 votes.  It turns out that Larry could have received anywhere between 3 and 6 votes.  To see this, first note that there are six different types of ballots submitted:

 Ballot LCM LMC CLM CML MLC MCL Voters a b c d e f

The table indicates that there were  a voters who ranked Larry first, Curly second, and Mo third.  The problem yields the following restrictions:

 (1) a + b + c + d + e + f = 20 because there are 20 citizens (2) a + b + e = 11 because 11 voters preferred Larry over Curly (3) a + c + d = 12 because 12 voters preferred Curly over Mo (4) d + e + f = 14 because 14 voters preferred Mo over Larry (5) a, b, c, d, e, f are whole numbers because they represent numbers of voters

With a little bit of algebra, restrictions (1)-(4) can be shown to be equivalent to

 (6) c = 6 – a – b by subtracting (4) from (1) (7) d = 6 + b by subtracting (4) from (2) (8) e = 11 – a – b by (2) (9) f = a – 3 by plugging in (7) & (8) into (3)

So, once  a and  b are chosen, the other variables are determined.  Since c ≥ 0, (6) implies a + b ≤ 6.  Since f ≥ 0 and b ≥ 0, (9) implies a + b ≥ a ≥ 3.  Since Larry receives a + b first-place votes, we have shown that Larry must receive between 3 and 6 votes.

The following table shows that Larry actually could have received anywhere between 3 and 6 votes.

 a + b a b c d e f 3 3 0 3 6 8 0 4 4 0 2 6 7 1 5 5 0 1 6 6 2 6 6 0 0 6 5 3

Note that there are other possible solutions, but they all give Larry between 3 and 6 votes.

October 16 Problem.  What does 1/2! + 2/3! + 3/4! + 4/5! + … equal? Note that the “…” means that the sum continues the previous pattern forever and n! = 1*2*…*n.

1.

Why by David Housman.

October 9 Problem. Consider a 4 inches by 4 inches square. Can five points be placed in the square so that every pair of points is at least three inches apart? If so, how? If not, why not?

Solution by Seth Unruh.  The Problem of the 4×4 square with points points all three inches apart is impossible. The first possible solution is one point at each corner and one in the middle, but this doesn’t work because the very middle is sqrt(32)/2 away from each point, and that is less than 3. So another solution would be to have one point in a corner and have points 3 inches away, and then another point 3 inches away. As seen by the attached diagram. the circles have a radius of three, as seen there is no point within the square which is outside of all the 3 inch circles for the other points. Thus the problem is impossible.

Solution by David Housman. Divide the 4-by-4 square into four 2-by-2 squares. Since there are 5 points and only 4 2-by-2 squares, there must be at least one 2-by-2 square containing at least 2 of the points. The furthest apart these two points could be are at opposite corners which are sqrt(8) < 3 apart.

October 2 Problem.  In a population in which 1 of 250 people were believed to have AIDS, a person obtained a positive first-stage test for the disease and committed suicide because he was sure that he must have AIDS. The test was believed to always obtain positive results if the person had AIDS but would also obtain a positive result for 4 percent of persons without AIDS (called a “false positive” in the medical literature). What was the probability that a person testing positive actually had AIDS?

Solution by Ginny Nunemaker.  Out of each 250 people, you would get 1 positive result for the person who actually has AIDS, and 9.96 positive results for the people who are false positive. (4% of the other 249 people)   So, out of 10.96 positives (the 1 that has AIDS, and the 9.96 false positives), only 1 would actually have the disease. Therefore the probability that a person testing positive actually has aids is 1 in 10.96 people, or approximately 9.1241%.  Since 96/100ths of a person would be a little strange to imagine, you could say that 100 people out of every 1,096 that test positive will actually have AIDS.

September 25 Problem. According to Benjamin Disraeli and Mark Twain, there are three kinds of lies: lies, damned lies, and statistics. The people of Linkton claim that their median income is \$1,000 higher than Rechtland’s median income. The people of Rechtland claim that their mean income is \$1,000 higher than Linkton’s mean income. Must one of the claims be a lie (if so explain why) or could both claims be true (if so explain how)?

Solution by Krista Nussbaum. Both claims could be true. Linkton could have a median \$1,000 dollars higher than Rechtland even if Rechtland’s mean is \$1,000 higher. This is because the mean is simply the average of the scores in a distribution, so there could be a lot of low scores, and then 1 super high outlier, therefore making the mean a lot higher, while the median would be lower, as most of the scores are lower. Also, Linkton could have the highest median, because even if it didn’t have any really high scores, the middle score in the distribution could be higher than the middle score in Rechtland’s distribution.
here are numbers.
Rechtland: 3,3,4,5,15 (in thousands) Median= 4, mean= 6
Linkton: 4,5,5,5,6 (in thousands) Median – 5, mean =5

September 18 Problem. David has 65 hens. If he had one more solid colored hen, then exactly one-third of his hens would be speckled.  From years of experience, David knows that one-half of the speckled hens will lay speckled eggs and that each hen and a half will lay an egg and a half in a day and a half. After how many full days will David have 4 dozen speckled eggs to sell?

Solution by Jonathan Savage. David has 65 hens, and adding one more unspeckled would make the speckleds a ratio of 1:2. So, 1 of every 3 in 66 is speckled, giving us 22 speckled hens. Only half of these hens will lay speckled eggs, so we then have 11 speckled egg laying hens. The problem gives us the information that 1.5 hens lays 1.5 eggs in a defined period of time. We can reduce this to a 1:1 ratio. In other words, one hen lays one egg in that period of time. So, now we have 11 hens laying 1 speckled egg in 1.5 days. We are required to produce four dozen eggs, or 4 x 12 = 48 eggs. Now, since we are required to round up to the next full day:
1.5 days = 11 eggs
3 days = 22 eggs
4.5 days = 33 eggs
6 days = 44 eggs
7 days = 44 + (11 x 2/3)* = 51 1/3 eggs. This surpasses our required amount.
*Represents the number of eggs produced in 1.5 days reduced to the number of eggs produced in 1 day, or 7 1/3 eggs. So, David will have four dozen eggs to sell after 7 full days.