Waldo is having a party and has 50 guests, among which is his brother Basil. Basil starts a rumor about Waldo; a person hearing this rumor for the first time will then tell another person chosen uniformly at random the rumor, with the exceptions that no one will tell the rumor to Waldo or to the person they heard it from. If a person who already knows the rumor hears it again, they will not tell it again. What's the probability that everyone, except Waldo, will hear the rumor before it stops propagating?
Puzzle #40: Fast Answer
I was sitting around with my friend Waldo, his nephew Spike, and Spike's friend Molly recently. I happened to have two tickets to a new movie in my pocket that I had just purchased, and I mentioned this and noted that there were two four-digit numbers on the tickets and that the sum of all 8 digits was 25. Waldo asked if any digit appeared more than twice out of the 8, which I answered, and then Spike asked if the sum of the digits of either ticket was equal to 13, which I answered also. Much to my surprise Molly immediately told me what the two numbers were! What were they?
Puzzle #41: Psychic Friends Network
A fair coin is flipped and then hidden. Molly has an 80% chance of guessing the state of the coin (heads or tails), Spike has a 70% chance, while poor Waldo only has a 10% chance of a correct guess. Basil wants to devise a scheme using all of these guesses to predict the state of the coin. What level of accuracy does the best scheme give?
Puzzle #42: Chess Logic
Eight players participated in the recent Cucumberland chess tournament; each player played all of the others exactly once. The winner of a game received 1 point and a loser 0; draws are allowed, giving each player 1/2 point. Now, it turned out that everyone received a different number of points. Furthermore, Molly, who came in second, earned as many points as the four bottom finishers put together. What was the result of the game between Waldo, who came in third, and Basil, who came in seventh?
At Basil's Burgers you can purchase Cheap Chicken Chunks in boxes holding 6, 9 or 20 Chunks. Therefore, you could purchase exactly 21 Chunks by purchasing two boxes of size 6 and one of size 9. Observe that there is no way to purchase exactly 17 Chunks; what is the largest number such that it is impossible to purchase that many Chunks exactly?
Solution:
The largest such number is 43. Note that the remainder upon dividing 20 by 3 is 2 and that 6 and 9 are multiples of 3, hence the only way to form a number that leaves a remainder of 1 upon dividing by 3 is to use at least two boxes of size 20, and it follows that it is impossible to make exactly 43. We can make 44 though 49, and so we can make all numbers greater than 43 by adding boxes of size 6.
Puzzle #32: Match Game
Make four hands of five cards, using three standard 52-card decks. What's the probability no hands have any matching cards?
Submitted by Kevin Schwartz.
Solution:
I don't have a nice solution yet, but I hope to have one by next issue (if one exists, that is).
Puzzle #33: Easy Temperature Conversion
When Waldo recently did a conversion of a positive integral Celsius temperature c=275 to its Fahrenheit equivalent f (which turned out to be 527), he noticed to his amazement that he could have simply moved the last digit of c to the front to obtain f. Doing some intense calculations he failed to discover the next largest such example. Does one exist, and if so, what is it?
Solution:
Let c = x_n * 10^{n-1} + ... + x_1 * 10^1 + x_0 with x_n > 0, then f = x_0 * 10^{n-1} + (c-x_0)/10. We also have that f = 9/5 * c + 32. Notice that in order for f to be integral c must be divisible by 5; this implies that x_0=5 since it cannot equal 0 (since as a number f>c). Our equation then becomes 9/5 * c + 32 = 5 * 10^{n-1} + (c-5)/10 ==> c = 5 * (10^n - 65)/17. Now it turns out that 10 is a primitive root modulo 17 (don't worry about what this means), and so it follows that c is integral if and only if n is of the form 16 * m + 3. When m=0 we get c=275; when m=1 we get the next highest such temperature, which is 5 * (10^19-65)/17 = 2941176470588235275.
Puzzle #34: Gift to Fire
Can you transform gift to fire, changing one letter at a time to make a new word? In how few steps? As an example, we can change cold to warm in four steps: cold cord word ward warm, and this is clearly best.
Solution:
The smallest number of steps I've found is four. One example is: gift lift life fife fire. This appears to be best using only common words. However, using obsolete words this may be reduced to three steps, the smallest possible number. One example is: gift girt gire fire. The word "gire" is in the Oxford English Dictionary as an obsolete 17th century variant of "gyre".
Puzzle #35: Multiple Birthdays
Mortimer and his grandson Waldo have the same birthday. For six consecutive birthdays Mortimer is an integral number of times as old as Waldo. How old is each at the sixth of these birthdays?
Solution:
Waldo was 6 and Mortimer was 66. To see that this is the only reasonable answer, we can apply the Chinese Remainder Theorem and get that if Waldo is w years old on the first such birthday and Mortimer m, m must be w plus an integral multiple of the least-common multiple of w through w+5. Since at least one of every six consective integers is divisible by 3, one by 4, and one by 5 it follows that 60 divides m-w; if w is one of 2 through 7 we have an additional factor of 7, implying m>420; if w>7 it follows from the fact that consecutive integers are relatively prime that m>12*13=156. Hence the only reasonable answer is that w is 1 and m is 61, leading to the answer above.
Puzzle #36: Birthday Ununiqueness
I was sitting around with my friend Waldo and his grandfather Mortimer last week, and the topic of birthday surprises came up. Mortimer mentioned that one of the greatest surprises that he has had involved his grandfather, who happens to have had the same birthday that Mortimer has. One year the family was celebrating this double birthday, and during the events Mortimer proudly mentioned to his grandfather that not only he had just turned as old as the last two digits of the year he was born in, but he was also a prime number of years old, and each of the two digits making up his age was also a prime. Mortimer was floored when the older man thought for a second, turned to him, and said that the same thing had just happened to him! What year did this occur, and how old had Mortimer and his grandfather just turned?
Solution:
It's clear that Mortimer had to have been born in the 1900s, and his grandfather in the 1800s. If Mortimer was born in 1900+x and his grandfather in 1800+y, then 1900+2x = 1800+2y (the year this happened), and so y = x+50. Now any odd prime number plus an odd number must be even and greater than 2 and so not a prime, hence Mortimer must have been 20-something. But 25, and 27 are not prime, so Mortimer must have been 23, and so his grandfather was 73 (which is indeed also prime). It follows the year was 1900+2*23 = 1946.
Puzzle #37: Factorial Over Easy?
Given that n! = 10888869450418352160768000000 find n (remember that n!=1*2*3* ... *n). There are several ways of doing this - the trick is to find the easiest.
Solution:
The number given ends in 6 zeroes, and so n must be between 25 and 29 inclusive. Discarding the six factors of 10 while multiplying the number from 1 to 29 and retaining only the last digit, we see that the last nonzero digit is 4 in 25!, 26!, and 28!; is 8 in 27!; and is 6 in 29!, so n must be 27.
Puzzle #38: Compuspeed
The Intel clock-doubled 486DX2-66 CPU chip operates by executing a certain fraction x of instructions totally on chip at a doubled rate (66 MHz), while the remaining 1-x are executed at the normal rate (33 MHz). It's observed that the 486DX2-66 is 76% faster than the 486DX-33 (which executes all instructions at 33 MHz). Given this and making some reasonable assumptions, estimate how much faster a clock-tripled 486DX3-99 (on chip 99 MHz; off chip 33 MHz) is than the 486DX2-66.
Solution:
The assumption I made was that the fraction x of instructions executed on-chip does not change in the clock-doubled 486DX2-66 CPU versus the clock-tripled 486DX3-99 CPU. Assuming a standard 486DX-33 CPU can execute 1 Unit (1 U) of instructions per second the 486DX2-66 would take {1\over 1.76} seconds to execute 1 U, and so x*U/(2*(U/s)) + (1-x)*U/(1*(U/s)) = s/1.76 implying that x/2 + 1 - x = .5682 and so x=.8636. It follows from the assumption that the 486DX3-99 executes one Unit in x/3 + (1-x) = .4243 seconds, so the 486DX3-99 is only about 33.9% faster than the 486DX2-66.
Such clock-tripled CPUs are indeed being sold, although for some reason they are called the 486DX4-100 (go figure). It should also be noted that these chips actually have a larger on-chip cache than the 486DX2-66 chips do (16k vs 8k), so the fraction of instructions executed on-chip by the 486DX4-100 is somewhat higher and thus they are faster than is predicted by above.