I'm thinking of a word that means someone who collects cheese labels. What is it?
Laclabphilist, as given in the excellent Dickson's Word Treasury.
This puzzle was seen floating around the internet.
Spike has a large bag of candies, each of which is one of 5 possible different flavors: apple, banana, cherry, dutch chocolate, and elderberry. Assuming he can fit up to 10 candies in his mouth at once, how many different flavors can he make? Note that 1 apple and 1 banana is the same flavor as 2 apples and 2 bananas (just a larger amount), but that 1 apple and 2 bananas is not the same as 2 apples and 1 banana. Also note that he has more than 10 candies of each flavor.
For a gold star, generalize the number of flavors and how many can be eaten at one time.
Let D(n) be the number of different flavors that can be made with n candies, and let E(n) be the number of different flavors that can be made with n candies, but not with any smaller number. The number we are seeking is F = E(1) + E(2) + ... + E(10). Note that D(n) = Sum E(d), where the sum is over all d such that d divides n. This gives us that D(6) = E(1) + E(2) + E(3) + E(6), D(7) = E(1) + E(7), D(8) = E(1) + E(2) + E(4) + E(8), D(9) = E(1) + E(3) + E(9), D(10) = E(1) + E(2) + E(5) + E(10). Thus, we can express F as D(6) + D(7) + D(8) + D(9) + D(10) - 4*E(1) - 2*E(2) - E(3) = D(6) + D(7) + D(8) + D(9) + D(10) - D(3) - 2*D(2) - D(1).
To find what D(n) is, imagine we have 4 sticks and n flavorless candies. Arrange the sticks and candies in a straight line; this can be done in (n+4) choose 4 = (n+4)!/(n!4!) ways. Note that each such arrangement corresponds to a different flavor by the simple mapping of letting all the candies to the left of the first stick be apple, the candies between the first and second sticks be banana, etc. This is a standard discrete math technique.
It follows that F = 14C4 + 13C4 + 12C4 + 11C4 + 10C4 - 7C4 - 2*6C4 - 5C4 = 2681. Generalizations follow immediately.
Last month we heard about some strange critters that live on Cucumber Island, lasamanders. There are some even stranger critters that live in Cucumberland proper, macheleons. As with the lasamanders there are three different types: azure, blue, and cream. When two macheleons of a different type touch, they will both take on the color of the third type, e.g. azure touching blue results in two cream macheleons. If there are 1994 azure, 1995 blue, and 1996 cream macheleons, is it possible for them to all become the same color by an appropriate series of meetings? For example, if there are 2 azure and 2 blue macheleons to start with, this is possible by an azure touching a blue, and then the remaining azure touching the remaining blue, leaving us with 4 cream macheleons.
For a gold star, give necessary and sufficient conditions on the starting numbers of each type of macheleon so that it's possible for them to all become the same color. For another gold star, if it's possible, what can be said about this color?
Let a, b, c be the number of azure, blue and cream respectively, then a necessary and sufficient condition that we may reduce to a single color is that some two of a,b,c be congruent mod 3. From this it follows that the desired reduction is impossible.
Necessity is obvious, since 2 = -1 (mod 3), thus the given operation (of touching and color change) maps each of a,b,c to an element equal to the (previous value)-1 (mod 3), thus two of a,b,c are equal mod 3 after the operation if and only if they were before. Since the desired final state consists of two 0 values, it follows that at least two of the three must have been equal mod 3 initially.
This is enough to solve our problem, but let's go further and prove sufficiency.
Sufficiency follows from induction. The truth is obvious if max(a,b,c)=1. Assume that the statement is true for max(a,b,c)=n, and note that we may assume that no two of a,b,c are equal, since if they are, we may easily reduce to the desired final state. Assume without loss of generality that a>b>c and that a=n+1. We then have that a=n+1, b ≤ n, and c ≤ n-2; note that c ≤ n-2 since a>b>c and we can't have a=n+1, b=n, c=n-1, since this does not have two of a,b,c congruent mod 3. Since a,b>0 we may pair one azure and one blue to form two cream, giving the state a-1, b-1, c+2, but now max(a-1,b-1,c+2) = n since a-1=n, b-1 ≤ n-1 and c+2 ≤ n. By the above observation two of a-1,b-1,c+2 are congruent mod 3 if and only if two of a,b,c are congruent mod 3, and since the statement is true for max(a,b,c)=n, the truth for n+1 now follows.
Another way of saying this is that if a,b,c are not all different mod 3, either two of a,b,c are equal, or we may reduce max(a,b,c). This is why induction works.
As for the final color, in the case where not all of a,b,c are equal mod 3 the final color must clearly be the one which was initially unique mod 3. If a,b,c are initially equal mod 3, then it can easily be proven by induction that the final color may be any unless two of a,b,c are initially 0.
My first three and last three are the same, and many people stay with me when they die. Who am I?
You're looking for an 11-letter word.
The word I was looking for was "underground".
Molly is playing the videogame Space Squeakers, and she finally loses her last turn after several hours (and after killing numerous invading space mice). She was rather surprised to discover that she had scored the maximum number of points she could have while averaging exactly 9975 points per turn. If in Space Squeakers you start with 3 turns and you earn an extra turn with every 10000 points you score (e.g. you earn an extra turn at 10000, another at 20000, another at 30000 etc.), what was Molly's final score?
Optional: For a gold star, describe the set of scores Molly could have achieved while still averaging 9975 points per turn.
Let T be the total number of turns Molly had (including the initial 3), and let S be her final score. Then by the rules of the game we have that S = 9975*T and T = 3 + floor[S/10000], where floor[x] is the greatest integer less than or equal to x. This gives that T = 3 + floor[9975*T/10000] = 3 + floor[(10000*T-25*T)/10000] = 3 + T + floor[-25*T/10000] = 3 + T - ceil[T/400], where ceil[x] is the smallest integer greater than or equal to x. We therefore get that ceil[T/400] = 3, and it follows that we must have 801 ≤ T ≤ 1200. Since Molly scored the maximum number of points, T = 1200, and so her final score was 9975*1200 = 11,970,000.
On Cucumber Island there are some strange critters, called lasamanders. There are three different types: alloy, bronze, and chrome. When two lasamanders of a different type meet, one will eat the other and take on the color of the third type, e.g. alloy meeting bronze results in a single chrome lasamander. If there are 1994 alloy, 1995 bronze, and 1996 chrome lasamanders, is it possible for them to reduce to a single lasamander by an appropriate series of meetings? For example, if there are 2 alloy and 1 bronze lasamanders to start with, this is possible by an alloy meeting the bronze and becoming chrome, then the remaining alloy meeting the chrome and becoming bronze.
Optional: For a gold star, give necessary and sufficient conditions on the starting numbers of each type of lasamander so that it's possible for them to reduce to a single lasamander. For another gold star, if it's possible, what can be said about the color of the last lasamander?
One ad hoc solution is to note that by repeatedly pairing alloy and bronze we may reduce to 0 alloy, 1 bronze, and 3990 chrome lasamanders, and we can clearly reduce to a single lasamander from this position. Another ad hoc solution is to note that the three successive meetings bronze-chrome, alloy-chrome, alloy-bronze result in reducing the number of each type by exactly one, so we may therefore reduce to 0 alloy, 1 bronze, and 2 chrome lasamanders, and the desired reduction is again obvious.
For the general case let a, b, c be the number of alloy, bronze and chrome lasamanders respectively, then a necessary and sufficient condition that we may reduce to a single lasamander is that not all of a,b,c are congruent mod 2 and that at least two of a,b,c are >0.
Aquatic creature cycles briefly to arise.
For those unfamiliar with the type of puzzle, it's of the form "This word, when you do this to it, becomes this other word". You're looking for a 7-letter word, and I'll take either the before or after.
MANATEE-EMANATE. The clue "aquatic creature" refers to "manatee", and the clue "arise" refers to "emanate". Finally, the clue "cycles briefly" refers to the fact that we cycle the letters in "manatee" by a small amount (in our case by 1 letter).
Spike is taking a series of exams, and it turns out that he'll have to score a 97 on the last one in order to average 90 for the entire series. But even if he scores as low as a 73, he'll still average an 87. How many exams were in the series?
Let S be the sum of all of his exam scores before the final, and let there be a total of n exams. Then we have (S+97)/n = (S+73)/n + 3, or 3n = 24, and so n = 8. In general, if a difference of x points on the final exam corresponds to a difference of y points in the overall average, there are x/y exams in total.
Waldo wants to draw an equilateral triangle (all sides the same length), and wants to do it as neatly as possible. To that end, he's using a large piece of graph paper made up of squares, and will have each edge go between two points on the graph paper.
Can such a neat triangle be drawn?
No. One way of seeing this is to note that by either Pick's Theorem or the determinant formula for the area of a triangle the area of an equilateral triangle with every vertex a lattice point must have the form n/2, for some integer n. Since the area of an equilateral triangle is also equal to sqrt(3)/4 * side^2, where side^2 is an integer if the vertices are lattice points, it would follow that if such a triangle existed then sqrt(3) would be rational, which is nonsense.
Alternately, we can assume that one of the vertices is at (0,0). Let the other two vertices be (a,b) and (c,d), and it then follows that a^2 + b^2 = c^2 + d^2 = 2(a*c + b*d). We may assume that at least one of a,b,c,d is odd (otherwise, keep dividing by 2 until this is the case). Assume that a is odd, then b must also be odd since a^2 + b^2 is even, and so a^2 + b^2 must be of the form 8k+2 for some integer k. Finally, it then follows from this that c and d must also be odd, and so 2(a*c + b*d) must be divisible by 4, contradicting the fact that it's of the form 8k+2.
"Irish wood is the best," Tom SAID.
I'm going to experiment with one off-beat verbal puzzle per column. In the problem above, you have to find a 6-letter word to replace SAID with that makes the above a nice Tom Swifty. A Tom Swifty is a phrase in which what Tom states is punningly related to how he says it; some examples:
"I love sugar in my coffee," Tom said sweetly.
"I'm dying," Tom croaked.
OPINED. This is a pun on both "Irish" (O') and "wood" (pine). Also note that Tom is, in fact, stating an opinion.
Donald Knuth, one of the most famous computer scientists in the world (and who was first published as a kid in Mad Magazine) believes that it's possible to make any positive integer by starting with a single 3 and then using some combination of the operations of factorial !, square-root sqrt(), and greatest integer []. Note that n! = 1*2*...*n (e.g. 6!=720), and that [x] is the greatest integer less than or equal to x (e.g. [3.14]=3). As an example, we can make 26 by [sqrt((3!)!)], since 3!=6, 6!=720, sqrt(720)=26.8, and [26.8]=26.
Show that it's possible to make 10.
As several people noted, I half gave the answer. Since [sqrt(26)] = 5 and [sqrt(5)!] = [sqrt(120)] = 10, we get one solution just by stringing everything together just so:
10 = [sqrt([sqrt([sqrt((3!)!)])]!)]
Note that there's an extra [] operation in there we don't really need.
Here's one from the USA Mathematical Talent Search; it's not easy, but it's certainly doable.
For a positive integer n, let P(n) be the product of the nonzero base 10 digits of n. Call n "prodigitious" if P(n) divides n. What is the maximum number of consecutive prodigititious positive integers n?
Hints: the answer is not 12; meditate on numbers ending in a 3, 6, or 9
Note that if any number n in our sequence ends in a 3, this implies 3 divides n. However, then n+10 also ends in a 3, yet is clearly not divisible by 3 (since 3 does not divide 10). The same trick works for numbers ending in 6 and 9, thus you can only have one of each in any consecutive sequence with the desired property. It follows that any such sequence has length at most 13, and indeed that it must go *0, *1, ..., *9, %0, %1, %2. Since I told you the answer was not 12, it has to be 13!
Consider sequences of the form (1)^n 000, (1)^n 001, ..., (1)^n 012, where for example (1)^3 000 = 111000. The freebies are the numbers ending in 00, 01, 02, 04, 05, 08, 10, 11, and 12. Let n be a multiple of 9, say 9*m; this gives us 03, 06, and 09 by the old trick of a number is divisible by 3 (9) if and only if the sum of its digits is divisible by 3 (9). The last number left is 07; we need that (10^{9m}-1)/9 is divisible by 7. This happens only if 10^{9m} = 1 (mod 7); by Fermat's little theorem this is ensured if 9*m is a multiple of 6 (and indeed, 10 is a primitive root modulo 7, so this is required). Let m=2k, then, so n=18k.
So we're done, the answer is 13, and one such sequence runs from 111,111,111,111,111,111,000 to 111,111,111,111,111,111,012.
The three of us made some bets:
Who am I?
Let Waldo start with w, Molly with m, and Spike with s. Following the above account we get the following progression of monies:
And so Waldo finished with w+m-s, Molly with 2*(m-w), and Spike with 2*(s+w-m). Since these must all be equal, we have three equations and three unknowns, so solve; this gives us that 4*m = 5*s and 3*s = 4*w. Now, if s = 1/2 this implies that w = 3/8 = 37.5 cents, an impossible amount of money to start with. If w = 1/2 this implies that s = 2/3, again an impossible amount to start with. Finally, if m = 1/2 this implies that s = 2/5 = 40 cents, w = 3/10 = 30 cents, which works. It follows that I am Molly.
Given that I*MENSA = ZZZZZZ and that each different letter stands for a different digit, what's Z?
"ZZZZZZ" = Z*111111 = Z*3*7*11*13*37, and so I must be either 3 or 7. But if I=3, then "MENSA" = Z*37037, giving "MENSA" = 37037 if Z=1 or "MENSA" = 74074 if Z=2; contradiction. Thus, I=7 and so "MENSA" = Z*15873. Since each letter must stand for a different digit, a quick check with a calculator gives that the only answer is Z=6.
Consider the sum:
ABC + DEF + GHI = JJJ
If different letter represent different digits, and there are no leading zeros, what does J represent?
Since there are no leading zeros, J must be 7, 8, or 9, since JJJ = ABC + DEF + GHI ≥ 100+240+350 = 690. Now, the remainder left after dividing any number by 9 is the same as the remainder left after dividing the sum of the digits of this number by 9. Next note that 0 + 1 + ... + 9 has a remainder of 0 after dividing by 9, and since JJJ has a remainder of 0, 3, or 6, it follows that J must be 9, since it's the only number from 7, 8 and 9 that leaves a remainder of 0, 3, or 6 if you remove it from the sum 0 + 1 + ... + 9.
No instructions. When you get the right answer, you'll know it.
HARD+HOKY=LOLY
HOKY+CIA=PYKYL
Just a little bit of lateral thinking required:
FOUR+FIVE=NINE
FIVE+TWO=SEVEN
Waldo mentioned the problem they had getting to the Cucumberland fair recently (see the Cheese and Fleas puzzle), and Mortimer recalled a similar occurence from the his days as a boy. It seems that three couples staying on an island wanted to cross the water using a boat that could only hold two people at a time. In those days, it was consider improper for a woman to be with a man who was not her husband unless her husband was also present. How many trips were required? Each way counts as one trip.
It can be done in 11 trips. Let W_i and H_i be husband and wife, then one such scheme is:
W_1 and W_2; W_1; W_1 and W_3; W_1; H_2 and H_3; W_2 and H_2; H_1 and H_2; W_3; W_2 and W_3; W_2; W_1 and W_2.
For each positive integer n, let A_n be the number of digits in the binary representation of n, and let B_n be the number of ones in the binary representation of n. What is (1/2)^(A_1+B_1) + (1/2)^(A_2+B_2) + (1/2)^(A_3+B_3) + ... ?
This is much easier than it looks.
Note that A_{2n} + B_{2n} = A_n + B_n + 1 and A_{2n+1} + B_{2n+1} = A_n + B_n + 2 and so letting S = (1/2)^(A_1+B_1) + (1/2)^(A_2+B_2) + ... we see that S = 1/2^2 + (1/2)^(A_2+B_2) + ... = 1/2^2 + (1/2 + 1/2^2)*((1/2)^(A_1+B_1) + ... and so S = 1/2^2 + S*(1/2 + 1/2^2). It follows that S/4 = 1/4 and therefore S=1.
Waldo is a bit absentminded, and the only way he can remember his own phone number is that if you divide it by its reverse, you get an integer greater than one. What's Waldo's phone number?
It could be either 879-9912 or 989-9901, so perhaps Waldo isn't quite as absentminded as all that.
Rameses wishes to build a great pyramid for his internment. The structure will have a square base and be solidly composed of cubical stone blocks. Each level of the pyramid contains one less block per side as the pyramid rises. Rameses has available an initial work force of 35,000 slaves. Each morning the available labor pool is divided into work crews of 17 slaves each. Any remainder that cannot form a full crew gets the day off but are available the following day. Each crew can lay one block of the pyramid each day. Unfortunately, the heat of the desert sun causes the death of one member of each crew each day. Work ceases on the project when it can be determined that there will be insufficient slaves available to raise the pyramid one more level. Each stone block measures 3 meters per side.
How many days will it take to construct Rameses' pyramid? How tall will it be? How many of the original slaves survive the construction?
This classic was printed in the October 1992 issue of PuzzleSIGns, the Mensa puzzle SIG publication. I don't know the original source.
Each block added results in the death of exactly one slave, and when there are less than 17 slaves no more blocks may be added, so the pyramid consists of at most 35000-16 = 34984 blocks. The first level needs 1 block, the next 4, and in general the ith level requires i^2 blocks, and so a pyramid n levels tall requires 1 + 2^2 + 3^2 + ... + n^2 = n(n+1)(2n+1)/6 blocks. This is between n^3/3 and (n+1)^3/3, and solving n^3/3 = 34984 we get that the maximum number of levels is either 46 or 47, with a quick check showing that 46 is correct. It follows that the pyramid is 46*3 = 138 meters tall, consists of 33511 blocks, and the number of slaves surviving the construction is 35000-33511=1489.
To compute the number of days required for construction, note that if there are m slaves on a given day the number of blocks added is between m/17 and m/17 - 16/17, and given k days it can be easily shown that the number of blocks added is between m/17 + (16/17)m/17 + ... + (16/17)^{k-1} m/17 and m/17 + (16/17)m/17 + ... + (16/17)^{k-1} m/17 - k*16/17; summing the geometric series we get m*(1-(16/17)^k), and so to find the number of days required we solve 35000*(1-(16/17)^k) = 33511 and get that (16/17)^k = .04254 or k=52.08. Since this came from an upper bound on the number of blocks, it's a lower bound on the number of days; thus, it'll take at least 53 days to build the pyramid. Checking the lower bound, we see that the number of blocks that can be added in 53 days is greater than 33542, and so the pyramid can indeed be built in 53 days.
Waldo, Basil, Molly, two wedges of cheddar, two wedges of mozzarella, and Rufus the dog are going to go to the annual Cucumberland cheese competition and dog show. Waldo's little sports car can only seat two objects at a time; Waldo, Basil and Molly can all drive the car. If Waldo is not around, Basil will eat the cheddar; if Basil is not around, Waldo will eat the mozzarella. If Molly is not around, Rufus will eat the cheese and bite Waldo and Basil. Can they all get to Cucumberland without anything bad happening? If so, what's the smallest number of trips needed (each way counts as one trip)?
Yes, they can do it, and the smallest number of trips is 17. There are essentially two ways of doing this in 17 steps. One way is:
Molly and Rufus; Molly; Molly and cheddar; Molly and Rufus; Waldo and cheddar; Waldo; Basil and Waldo; Basil; Molly and Rufus; Waldo; Basil and Waldo; Basil; Basil and mozzarella; Molly and Rufus; Molly and mozzarella; Molly; Molly and Rufus.
The other method is to do this in reverse. Another way of looking at it is that we're swapping cheddar and mozzarella, Basil and Waldo, since from the point of view of the other restrictions this changes nothing.
Waldo is having a party and has 50 guests, among whom 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?
Let person 1 be the initiator, and person m>1 be the mth person to hear the rumor. Then the probability that person m will give the rumor to a person who hasn't heard it yet is (48-(m-2))/48, since there are 48 people who can hear the rumor, and of these m-2 have already heard it. Thus the probability that the rumor will propagate to everyone is the product (50-2)/48 * (50-3)/48 * ... * (50-49)/48, which equals 48!/48^48.
The Cucumberland grocery has six cheese wedges of different sizes, weighing 15, 16, 18, 19, 20, and 31 pounds. Five wedges are cheddar and only one wedge is mozzarella. Waldo bought two wedges of cheddar, and Basil also bought cheddar, but twice as much by weight as Waldo. How much does the mozzarella wedge weigh?
Waldo must have bought at least 31 pounds of cheese, and so Basil must have bought three wedges since the two heaviest wedges together only total 51 pounds. Since Basil bought twice what Waldo bought, the total amount of cheese bought by the two together equals 3 times that purchased by Waldo alone. Now, the total weight of all the wedges at the grocery divided by 3 leaves a remainder of 2, so the mozzarella wedge must leave a remainder of 2 when divided by 3. But the only such wedge is the 20 pound wedge, thus the mozzarella wedge must have weighed 20 pounds.
Waldo and Basil were recently invited to a party attended by four other pairs of siblings, for a total of ten people. During the party various handshakes took place, but no person shook their own hand or the hand of their sibling. At the end of the party Waldo asked each person, including Basil, how many different people they shook hands with, and was surprised to note that every number was different! How many hands did Basil shake?
Since each person shook a different number of hands, the totals must range from 0 to 8. Now the person who shook 8 hands must have shaken the hand of everyone except their sibling, so their sibling must be the one who shook 0 hands. It follows that the persons who shook 1 and 7 hands must be siblings, since we must have that 1 only shook with 8, 7 with everyone but 0 and 1. In a similar fashion the persons who shook 2 and 6 hands must be siblings and the persons who shook 3 and 5 hands must be siblings. Since the only number left is 4, it follows that Basil must have shaken 4 hands.
Prince Waldo went to fight a 3-headed, 3-tailed dragon. He has a magic sword that can, in one stroke, chop off either one head, two heads, one tail, or two tails. This dragon is of a type related to the hydra; if one head is chopped off, a new head grows. In place of one tail, two new tails grow; in place of two tails, one new head grows; if two heads are chopped off, nothing grows. What is the smallest number of strokes required to chop off all the dragon's heads and tails, thus killing it?
All of the tails must be converted to heads in such a way that the dragon is left with an even number of heads. The way to do this in the smallest number of strokes would be to first convert the 3 tails to 6 tails by using 3 1-tail strokes, then converting the 6 tails to 3 heads via 3 2-tail strokes. Finally, the now 6-headed dragon can be killed with 3 2-head strokes, for a total of 9 strokes.
Another solution is to let the number of 1-head strokes be a, 1-tail strokes be b, 2-head strokes be c, and 2-tail strokes be d. Then the number of heads h after these strokes is h = 3-2c+d; tails t is t = 3+b-2d. If both are 0 we have that 2c = d+3 and 2d = b+3. It's clear that d must be odd, so let d = 2m+1, then c = m+2 and b = 4m-1. The least number of strokes occurs when a=0 and m is the smallest integer that makes b, c, d non-negative, and so m=1 implies b=3, c=3, d=3, giving us the same answer as before.
After graduating from college, you have taken an important managing position in the prestigious financial firm of Waldo & Spike. You are responsible for all the decisions concerning takeover bids, and your immediate concern is whether to try a takeover of Basil's Brokerage. There is no doubt that you will be successful if you are the first to bid and that this will be profitable for the firm and you in the long run. However, you know that there exist another n financial firms, similar to Waldo & Spike, that are also considering the possibility. Although you are likely to be the first one to move, you know that just after a takeover there is a lot of adjustment that needs to be done. In fact, for a period of time following any takeover the successful firm becomes a prime candidate for a takeover which will cost the job of whoever is responsible for takeovers (in other words, you). Among all financial firms it is common knowledge that the managers responsible for takeovers are rational and intelligent. What is your best response?
Assume the takeover is wise for n. The takeover is then unwise for n+1, as the other companies now find themselves in the same situation as you for n. If the decision is unwise for n, by similar reasoning it is wise to takeover Basil's Brokerage for n+1. Now note that for n=1 the takeover decision is clearly unwise, hence by induction you should takeover Basil's Brokerage if and only if n is even.
We were sitting around recently discussing birthdays (as we often do), and Mortimer mentioned that once on his father's birthday in 1937 someone (who didn't know it was his birthday) asked his father when he was born. He told him that he turned x years old in the year x^2, and that if you added his current age to the number of the month he was born in, it equaled the day of the month on which he was born. When was Mortimer's father born?
Mortimer's father was x year old in the year x^2, and we know he was alive in 1937. Now x can't be 43 since 43^2 is 1849 (making him 131 in 1937) and 45^2 is 2025 (wouldn't be born until 1980). So he must have turned 44 in 44^2 = 1936, so he was born in 1892. He turned 45 in 1937, so if d is the day and m is the month he was born on/in, we have that 45 + m = d^2. Since 1 le m le 12 gives a range for d^2 of 46 to 57, m must be 4, since the only perfect square in this range is 49. It follows that d is 7 and so he was born on April 7, 1892.
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?
An easy way of solving this is to note that if each participant beat each person who finished below them, this would fit the given criteria. Since Waldo finished above Basil, if the puzzle has a definite answer Waldo must have beaten Basil.
To show this more rigorously, let x_i be the score of the person who finished in ith place. Then x_2 ≤ 6, since clearly x_2 cannot be 7, and if x_2 = 6.5 this would imply x_1 = 7, but then person 1 would have beaten everyone, hence x_2 could not be 6.5. Now players 5, 6, 7 and 8 played exactly 6 games among themselves, hence x_5 + x_6 + x_7 + x_8 ≥ 6. Since we are given that x_2 = x_5 + x_6 + x_7 + x_8, we conclude that x_2 = 6. To finish, note that since the sum x_5 + x_6 + x_7 + x_8 is exactly 6, neither player 5, 6, 7 or 8 could have beaten or tied any of players 1, 2, 3 or 4, since then this sum would be greater than 6. It now follows that Waldo beat Basil.
Mortimer recently had another birthday. When someone mentioned that he was getting up there in years, he replied that he was actually quite young. Indeed, he pointed out, he is the youngest age such that the sum of the divisors of his age, not including the age itself, exceeded his age, yet the sum of no subset of these divisors equaled his age. How old had Mortimer just turned?
Clearly this number cannot be prime, nor can it be a product of two primes. To see this latter case, let these primes be p and q; if p=q the fact that the sum of factors cannot exceed the number is easy to see. If p and q are distinct, with p>q, we have that p(q-1) ≥ p ==> pq ≥ p+p ≥ p+q+1 and the same conclusion follows. Looking at the first few examples we see that there does exist such a number with three distinct prime factors, 70 = 2*5*7, since the sum of its factors not including itself is 1+2+5+7+10+14+35=74, but no subset of these factors sums to 70. We still need to check those numbers smaller than 70 that have 3 or more prime factors (not necessarily distinct), but that's easily done.
Numbers that have these properties are known to mathematicians as weird numbers. Weird numbers are rare, and the next smallest are 836, 4030, 5830, and 7192.
Define the sequence f(n) in the following manner:
f(0) = 0, f(n) = n - f(f(n-1))
What is f(1000000000000)?
Clearly f(n) is always less than n, and checking the first few values by hand it appears to be increasing, with f(n+1) equaling either f(n) or f(n)+1. A reasonable guess is that f(n) is approximately c*n, for some constant c. To find what c is if this is the case, consider f(n)/n = 1 - f(f(n-1)))/n. In the limit as n goes to infinity, the left side becomes c, and the right side becomes 1 - c^2. Since one must equal the other c=1-c^2, and rejecting the negative root we have that c is (-1+sqrt(5))/2. Examining the values of c*n rapidly leads to the conclusion that f(n)= [c*(n+1)], where [x] is the smallest integer less than or equal to x.
To prove this, note that if x>0 is not an integer, then [-x] = -[x] - 1. Now n - [c*[c*n] + c] = n + [-c*[c*n] - c] + 1 = n + 1 + [ -c^2*n - c*e - c ], where e = c*n - [c*n]. Since c^2 = 1-c, this equals n + 1 + [-n - c*n - c*e - c] = 1 + [c*n - c*e - c] = [c*n + c] + [f - c*e - 2c + 1], where f = c*n + c - [c*n + c]. Now if e+c<1, f=e+c; if e+c ≥ 1, f=e+c-1. It follows easily from this that [f - c*e - 2c + 1] is always 0. Since f(0) = 1 = [c*0 + c], the truth of our hypothesis follows, and it also follows that f(1000000000000) = 618033988750.
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?
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.
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?
What Molly realized was that the tickets were consecutively numbered. If the tickets were numbered abcd and abc(d+1) and my answer to Spike's question had been yes, the only conclusion Molly could have reached would have been that a+b+c+d=12, and regardless of my answer to Waldo's question there would not have been a unique solution. So my answer to Spike's question must have been no, and it follows that the tickets could not have been numbered in this manner. If the numbers were abc9 and ab(c+1)0 we'd have 2a+2b+2c+10=25, and so 2(a+b+c)=15, which is impossible. If the numbers were ab99 and a(b+1)00 we'd have 2a+2b+19=25 or a+b=3, leading to the four possibilities 0399 and 0400, 1299 and 1300, 2199 and 2200, 3099 and 3100. Of these, three of them would have had me answer "yes" to Waldo's question, and only the pair 1299 and 1300 would have had me answer "no". It follows that these were my ticket numbers.
This puzzle was based on one appearing in London's Sunday Times.
When Mortimer turned x years old he noticed to his surprise that between them x^2 and x^3 included all the digits from 0 to 9, with none repeated. How old was he?
Mortimer was 69; 69^2 = 4761 and 69^3 = 328509. The number of ages to check can be reduced by noticing that if x is Mortimer's age, x^2 must have 4 digits and x^3 must have 6, and so we have 10^5 ≤ x^3 < 10^6, giving that 47 ≤ x < 100. Furthermore, x can't end in a 0, 1, 5, or 6, as then x^2 and x^3 would have the same last digit. The number of cases to check can be reduced still further, but there are only 33 cases left so there isn't much of a point.
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?
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 implies 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.