So You Think You've Got Problems
Page 16
However, no number fits the gap, so we can eliminate TWO as wrong. Let’s try FOUR:
The lowest number that works is TEN:
We complete the sequence by continuing in this vein.
85 SEXY LEXY
One billion: it has ten letters, and 1,000,000,000 has ten digits.
86 LETTERS IN A BOX
87 WONDERFUL WORDS
[1] Wholesome.
[2] It is the longest English word that reverses to make a different word: in this case, desserts.
[3] Natal, Nice and Reading are all places.
[4] ‘Yes’ becomes ‘Ayes.’
[5] Short.
[6] ‘Nowhere’ becomes ‘now here’.
[7] ‘Queue’, which has the same pronunciation as ‘Q’. Or ‘aitch’, which has the same pronunciation as ‘H’.
[8] Because end does indeed begin with an ‘e’.
[9] The adjectives are synonymous with mercurial, venereal, earthy, martial, jovial and saturnine, which each come from the names of the planets, in order of their distance from the Sun: Mercury, Venus, Earth, Mars, Jupiter and Saturn.
[10] A post box.
88 LIFE SENTENCES
[1] Buffalo buffalo Buffalo buffalo buffalo buffalo Buffalo buffalo.
The word ‘buffalo’ has three meanings: as a city, as an animal and as a verb. One way to make the sentence less confusing is to clearly distinguish between these meanings by rewriting the city name as ‘from the city of Buffalo’, and to substitute the animal for ‘bison’ and the verb for ‘intimidate’. So we get:
Buffalo buffalo (i.e., bison from the city of Buffalo) Buffalo buffalo buffalo (i.e., the ones that are intimidated by bison from the city of Buffalo) buffalo Buffalo buffalo (i.e., intimidate bison from the city of Buffalo).
In other words, the bison from Buffalo that are intimidated by other bison from Buffalo, themselves intimidate bison from Buffalo.
[2] John, where James had had ‘had’, had had ‘had had’; ‘had had’ had had the teacher’s approval.
Here’s a context in which this makes sense. A teacher asked John and James to write an English sentence in the pluperfect tense. John wrote the sentence ‘The woman had had a baby.’ James wrote ‘The woman had a baby.’
[3] Sunday.
The day after tomorrow is in two days’ time. So when the day after tomorrow is yesterday, this date – ‘today’ – will be in three days’ time.
Likewise, the day before yesterday is two days ago. So when the day before yesterday is tomorrow, this date – ‘today’ – is three days ago.
If these two dates – three days ago, and in three days’ time – are equidistant from Sunday, then today must be Sunday.
89 IN THE BEGINNING (AND THE MIDDLE AND THE END) WAS THE WORD
The word is ‘only’.
90 LOOKING AT LETTERS
[1] It’s an E sideways.
[2] It’s the alphabet with a twist! Or with successive quarter-turns. Listing the letters backwards from Z, the Z has been rotated 90 degrees (either clockwise or anticlockwise). The Y has not been rotated, the X has been rotated by 90 degrees (either clockwise or anticlockwise). The fourth letter is a W that has been rotated 180 degrees. The next symbol would be a V turned by 90 degrees, which is either < or >.
[3] Turn the page sideways and you will read ONION, NOON, ZOO and NUN.
91 A MATTER OF REFLECTION
The answer is HID, which is the only word in the text made from letters that look the same when reflected horizontally.
92 THE BLANK COLUMN
The phenomenon occurs for all sentences, provided the text is aligned with the left margin and words are not broken on the right.
Let’s consider the sentence in the question: ‘So you think you’ve got problems?’ In the second line, which starts with ‘got problems?’, these words are immediately followed by ‘So you think you’ve.’ In other words, the first six words of the line are precisely the six words in ‘So you think you’ve got problems,’ just not in the correct order.
Likewise, in the third line, which starts ‘think you’ve got problems?’, the first six words are also the six words in ‘So you think you’ve got problems,’ just not in the correct order. In fact, whichever word starts the line, the first six words will always be the six words in ‘So you think you’ve got problems?’ If the sentence starts with the same six words, it follows that the sixth word will always end at the same position along, and therefore immediately after the sixth word there will be a blank space.
We can generalise this to any sentence S with n words. Whatever word begins the line, the first n words of that line are precisely the words in S. It follows that the nth word must always end in the same position, so immediately after the nth word (or the punctuation point after that word) is a space.
93 WELCOME TO THE FOLD
94 MY FIRST AMBIGRAM
Scott Kim gives these as possible solutions:
95 BOXED PROVERBS
[1] Look before you leap.
[2] A stitch in time saves nine.
[3] A rolling stone gathers no moss.
[4] Silence is golden.
[5] Beggars can’t be choosers.
[6] All’s well that ends well.
[7] Beauty is only skin deep.
[8] All roads lead to Rome.
[9] Better safe than sorry.
[10] You can’t judge a book by its cover.
96 NMRCL ABBRVTNS
[1] Winter, spring, summer and autumn are the 4 seasons.
[2] Africa, Asia, Australia, Antarctica, Europe, North America and South America are the 7 continents.
[3] Sneezy, Grumpy, Dopey, Bashful, Sleepy, Doc and Happy are the 7 dwarfs.
[4] The alphabet has 26 letters.
[5] A guitar has 6 strings.
[6] There are 52 cards in a deck.
[7] A spider has 8 legs.
[8] A chessboard has 64 squares.
[9] A golf course has 18 holes.
[10] A 1 followed by 6 zeros is a million.
97 THE NAME OF THE FATHER
98 TELLING THE TIME IN TALLINN
From the clock faces we can deduce straight away that üks is 1, kaks is 2, and viis is probably 5. (Since it’s likely that minutit means ‘minute’ and läbi means ‘past’.) The words veerand and kolmveerand seem to refer to a quarter hour and three quarters of an hour, so kolm is probably 3.
What’s confusing now is that the expression for 1.15 has kaks, the word for 2, in it, and the expression for 10.45 has üksteist in it, which looks like it must be 11. (Because the only other similar expression in the question is kaksteist, which would be 12.)
The deductive leap is to realise that, in Estonian, a ‘quarter past the hour’ is instead described as ‘a quarter of an hour on the way to the next hour,’ ‘half past the hour’ is described as ‘half an hour on the way to the next hour’ and ‘three quarters past the hour’ is described as ‘three quarters of an hour on the way to the next hour.’ In seeing that both pool neli and veerand neli are expressions, we can deduce that neli is 4, which means that the only number left, üheksa, must be 9. So the answers are:
Estonian to numbers:
[1] 9.25
[2] 11.45
[3] 2.30
[4] 3.15
[5] 6.35
Numbers to Estonian:
[1] Veerand viis.
[2] Kolmveerand üheksa.
[3] Pool kaksteist.
[4] Viis minutit seitse läbi.
[5] Pool üks.
99 COUNTING IN THE RAINFOREST
[1] aroke
[2] mẽña
[3] mẽña go aroke
[4] mẽña go mẽña
[5] ãẽmãẽmpoke
[6] ãẽmãẽmpoke go aroke
[7] ãẽmãẽmpoke go mẽña
[8] mẽña mẽña mẽña mẽña
[9] ãẽmãẽmpoke mẽña go mẽña
[10] tipãẽmpoke
From [2], there are tw
o numbers such that the sum of their squares is at most 10. These two numbers must be either of 1, 2 or 3, so we can infer that aroke and mẽña are 1, 2 or 3, and that ãẽmãẽmpoke is 5 or 10.
From [4] we can see that mẽña cannot be 1, so ãẽmãẽmpoke must be 5, since if it was 10 its product with mẽña would be greater than 10. If ãẽmãẽmpoke is 5, aroke and mẽña are either 1 or 2. But since mẽña is not 1, it must be 2, which means that aroke is 1; thanks to [4] tipãẽmpoke is 10.
We still have to account for 3, 4, 6, 7, 8 and 9. From [3], the number which is squared, ãẽmãẽmpoke go aroke, cannot be 3, since the only two numbers that multiply to equal 9 are 1 and 9, and 1 is already taken. Likewise, we can eliminate 4, 7, 8 and 9, so ãẽmãẽmpoke go aroke must be 6, and the two numbers on the other side of the equation are 4 and 9 (since 4 x 9 = 36). The expression on the left is mẽña go mẽña, and the one on the right is ãẽmãẽmpoke mẽña go mẽña – in other words, the same expression but including the word for 5. So it makes sense for mẽña go mẽña to be 4 and for ãẽmãẽmpoke mẽña go mẽña to be 9.
Now to [1]: mẽña mẽña mẽña mẽña = 12 – 4 = 8. The missing numbers now are 3 and 7. We build 3 (which is 2 + 1) by analogy to 6 (which is 5 + 1) and we build 7 (which is 5 + 2) by analogy to 4 (which is 2 + 2).
The solution reveals an underlying structure that is a mixture of base two and base five.
100 CHEMISTRY LESSON
This question requires intelligent guesswork as well as pure logic.
The names have four possible prefixes: hept-, prop-, but- and eth-; and three possible suffixes: -ane, -ene and -yne.
Likewise, there are four types of carbon: C2, C3, C4 and C7. A sensible guess is that the carbon numbers correspond to the prefixes, and that C7 is hept-, since hept- is a prefix that in other contexts such as ‘h’ and ‘t’ means 7. Two formulae contain C3 and two C4, and likewise two names contain prop- and two contain but-, but only one formula contains C2 and only one name has eth-, so we can deduce that C2 must be eth-.
So far we have:
Heptene: C7H14
Ethyne: C2H2
There are five types of hydrogen: H2, H4, H6, H8 and H14, but only three suffixes, so suffixes do not correspond to the type of hydrogen.
What else could the suffixes refer to? Let’s consider the ratios between the numbers. With heptene, the H number is double the C number. There is only one other -ene, and only one other formula in which the H number is double the C number. So let’s go with the idea that -ene means H, which is double C. So:
Butene: C4H8
Which leaves us with one remaining C4, which must be the other but-.
Butyne: C4H6
Now we seem to have run into a problem. If the suffix is based on the ratio between C and H, why do ethyne (C2H2) and butyne (C4H6) have the same suffix but different ratios? The insight here is to realise that H-numbers are always even, so the ratio might have something to do with being half the H-number. This would give ethyne a C:(H⁄2) ratio of 2:1, and butyne a C:(H⁄2) of 4:3. In both cases C = (H⁄2) + 1, which suggests that -yne occurs when C = (H⁄2) + 1.
The two remaining formulae have a C:(H⁄2) ratio of 3:2 and 3:4, which would make the former propyne and the latter, by elimination, propane. So:
Propyne: C3H4
Propane: C3H8
Tasty teasers
Bongard bafflers
1) Left: sets of three colinear points. Right: no sets of three colinear points.
2) Left: circles on different curves. Right: circles on the same curves.
3) Left: more circles inside than outside. Right: fewer circles inside than outside.
4) Left: black region widens in the direction of the opposite side of the shape. Right: black region narrows in the direction of the opposite side of the shape.
Sleepless nights and sibling rivalries
PROBABILITY PROBLEMS
101 BETTER THAN HALF A CHANCE
Place a single dark chocolate cookie in one jar, and the rest in the other jar. With this arrangement, if you open one of the jars you will have a 100 per cent chance of choosing a dark chocolate cookie, and if you open the other you will have a 49⁄99 = 49.49 per cent chance. So on average you have almost a 75 per cent chance of getting your preferred flavour.
102 SINGLE WHITE PEBBLE
If the urn contains an even number of pebbles the chances of you or your friend picking the white pebble are equal. Going first conveys no advantage. You might as well separate the pebbles into two equally sized groups. The white pebble has a 50 per cent chance of being in either group.
In fact, the probability of picking the white pebble at any stage is the same. If there are n pebbles in total, you have a 1⁄n chance of getting the pebble on your first pick. The chances of getting it on your second pick are the chances of not getting it on the first pick, which are (n–1)⁄n, multiplied by the chances of getting it on the second pick, which are 1⁄(n –1). This gives you a 1⁄n chance.
If the urn contains an odd number of pebbles, it’s worth going first because you get one more attempt.
103 THE JOY OF SOCKS
Four socks in total: two red and two blue. The minimum number of socks you need to take out to ensure you have either two of the same or two of different colours is three.
104 LOOSE CHANGE
If you take 20 coins out of your pocket and at least one is a 10p piece, this implies that there must be at least seven 10p pieces among the 26 coins. (If there were six or fewer 10p pieces you would not be guaranteed to get one among the 20 coins you take out.) Likewise, if you take out 20 coins and at least two are 20p pieces, you must have at least eight 20p pieces in your pocket, and if you take 20 coins out and at least five are 50p coins, you must have at least eleven 50p pieces.
If you have at least seven 10p, at least eight 20p and at least eleven 50p pieces, you must have exactly those amounts, since 7 + 8 + 11 = 26. So the total money in your pocket is 11 x 50p + 8 x 20p + 7 x 10p = £5.50 + £1.60 + 70p = £7.80.
105 THE SACK OF SPUDS
There are 11 potatoes in the sack, which means that there are 211 = 2,048 ways that we can take a number of potatoes from the sack. We could take no potatoes. Or we could take one potato, two potatoes, three potatoes, … – stop me now before I break into song.
The 2,048 combinations of potatoes that we can take from the sack will have weights that vary between 0g and 2,000g. Since there are more possible combinations than there are weights, there must be at least two combinations of potatoes that weigh the same number of grammes (rounding to the nearest gramme). Let’s call these two combinations A and B.
If A and B do not have any potatoes in common, we can remove each of these combinations from the sack, and the weight of the pile made from A will differ from the weight of the pile made from B by less than 1g.
If A and B do have potatoes in common, we can simply remove these common potatoes, and what’s remaining in pile A and what’s remaining in pile B will still have weights that differ by less than 1g.
106 THE BAGS OF SWEETS
Fifteen sweets. Place a sweet in the first bag, then place this bag, with another sweet, in the second bag, then place these two bags, with another sweet, into the third bag, and so on. The fifteenth bag will contain all 15 sweets and all the other bags too. Stashed in this way, each bag will contain a different number of sweets (and bags).
107 A STRATEGY FOR THE DISPLACEMENT OF IMPROPER THOUGHTS
The chance that the ball in the bag will be white is 2⁄3, or 66.7 per cent.
We solve this problem by considering equally likely outcomes.
The ball in the bag is either black or white. When a white ball is placed in the bag, there are two equally likely possibilities: either the bag now has a black and a white ball, or it now has two white balls. When a ball is taken out of the bag at random, there are now four equally likely possibilities, as described in this table. (When there are two white balls, one is called white1, and t
he other is white2.)
Balls in the bag Ball taken out Ball still in bag
black, white black white
black, white white black
white1, white2 white1 white2
white1, white2 white2 white1
There are three possibilities in which a white ball has been removed, and in two of them the ball inside the bag is white. So the chance of a white ball being in the bag is 2 in 3, or 66.7 per cent.
108 BERTRAND’S BOX PARADOX
The chance that the other counter in the box is black is 2⁄3, or 66.7 per cent.
Like the previous problem, we need to consider all equally likely outcomes.
When you open a box at random and chose a counter at random, every counter has an equal chance of being chosen. If the counter is black, you have chosen one of the three black counters. The crucial insight here is that each black counter is an equally likely choice. I’ve labelled the black counters A, B and C.
If you removed A, the other counter in the same box is black.
If you removed B, the other counter in the same box is black.