0588235294117647 × 16 = 9411764705882352
For my second question:
0588235294117647 × 17 = 9999999999999999
The source of this remarkable number is the decimal expansion of the fraction 1/17, which is
0.0588235294117647 0588235294117647
0588235294117647 . . .
repeating indefinitely.
Which is Bigger?
By direct calculation, eπ = 23.1407 whereas πe = 22.4592. So eπ > πe.
Actually, there’s a more general result: ex ≥ xe for any number x ≥ 0, and equality holds if and only if x = e. So, not only is eπ > πe, but e2 > 2e, e3 > 3e, e4 > 4e, e√2 > (√2)e, and indeed e999 > 999e. The simplest proof uses calculus, and here it is for anyone who wants to see the details. It also helps to explain why eπ and πe are so close together.
Let y = xee-x, where x ≥ 0. We find the stationary points (maxima, minima, and the like) by setting dy/dx = 0. Now
which is zero at x = 0 and x = e, and nowhere else. The value of y at x = 0 is y = 0, which is clearly a minimum; the value at x = e is y = 1. This is in fact a maximum. To see why, calculate the second derivative
at x = e. This works out as -1, which is negative, so x = e is a maximum, and the maximum value of y is 1.
Therefore xee-x ≤ 1 for all x ≥ 0, with equality only at the unique maximum x = e. Multiply both sides by ex, to get
xe ≤ ex
for all x ≥ 0, with equality only when x = e. Done!
The graph of the function y = xee-x has a single peak at x = e, and tails off towards zero as x increases to infinity.
Graph of y = xee-x.
This helps to explain why eπ and πe are so close together for it not to be immediately obvious which is bigger. The graph also shows that if a number x is reasonably close to e, then xee-x is close to 1, so xe is close to ex. For example, if x lies between 1.8 and 3.9, then xe is at least 0.8ex. In particular, this holds for x = π.
Colorado Smith and the Solar Temple
The division shown solves the puzzle. So does its reflection in the diagonal.
Four regions of the same shape, each containing a sun-disc.
Why Can’t I Add Fractions Like I Multiply Them?
The short response is that we can’t add fractions that way because we don’t get the right answer! Since is nearly , and so is , then when we add them the result must be at least . But is less than because half of 12 is 6. The error is even more glaring when we try it on + , because
makes no sense: since it tells us that .
All very well, but why does the multiplication rule work, and what should we use for addition instead?
The easy way to see why the rules differ - and what they should be - is to use pictures. Here’s a picture for × .
Multiplying fractions.
The vertical bar shows a line of five equal pieces, with two of them shaded grey. That represents : two parts out of five. Similarly the horizontal bar represents . The rectangles represent multiplication, because the area of a rectangle is what you get when you multiply the two sides. The big rectangle contains 5 × 7 = 35 squares. The shaded one contains 2 × 3 = 6 squares. So the shaded rectangle is of the big one.
When it comes to addition, the corresponding picture looks like this:
Adding fractions.
We get of the big rectangle by taking the top two rows out of five, and by taking the left-hand three columns out of seven. These regions are shown in the left-hand picture, with different shading, and they overlap. To count how many squares there are altogether, we either count the overlapping squares twice, or make an extra copy as in the right-hand picture. Either way, we get 29 squares out of 35, so the sum must be
To see how the 29 relates to the original numbers, just count the squares in the top two rows, 2 × 7, and add those in the left-hand three columns, 3 × 5. Then 2 × 7 + 3 × 5 = 29. So the addition rule is
This is where the usual recipe ‘put both fractions over the same denominator’ comes from.
Pooling Resources
It was a bad idea. Separately, Christine would make £150 and Daphne £100, a total of £250. Together, their total income would be £240 - which is smaller.
Both pairs of ladies are making an unwarranted assumption; it happens to work in favour of the first pair and against the second pair. The assumption is that the way to combine prices of a for £b and c for £d is to add the numbers, getting a + c for £(b + d). This boils down to trying to add the corresponding fractions using the rule
and we have seen with the previous puzzles that this doesn’t work. Sometimes it is an underestimate, sometimes an overestimate. It is correct when the two fractions concerned are the same.
Welcome to the Rep-Tile House
This shape is rep-9.
Cooking on a Torus
A torus can be represented as a rectangle with opposite faces identified - that is, ‘wrapped round’ so that anything that disappears off one edge reappears at the opposite edge. A Möbius band can be drawn as a rectangle with the left- and right-hand edges identified, but with a half twist. Drawn that way, here are possible solutions. Remember, if you draw it on a Möbius band made from paper, then the lines are deemed to ‘soak through’ the sheet of paper.
Connecting utilities on a torus . . .
... and a Möbius band.
The Ham Sandwich Theorem
Lots of examples prove that in general you can’t bisect three shapes with a single straight line. Here’s one with three circles. It’s easy to show that the only line bisecting the lower two is the one illustrated. But this doesn’t bisect the third one.
Only one line bisects the two lower circles, and this line does not bisect the top one.
In three dimensions, exactly the same idea works with four spheres. The centres of three of them lie on some plane, and provided those centres are not in a straight line, there is exactly one such plane. Now put the centre of the fourth sphere at a point that is not on the plane.
Cricket on Grumpius
The Grumpians are septimists, and use base-7 arithmetic. In their system, 100 works out as
1 × 72 + 0 × 7 + 0 × 1 = 49
So they get very excited instead of being disappointed: the batsthing with decimal 49 runs has just scored a Grumpian century!
The Missing Piece
Innumeratus’s solution.
Well, it does look pretty convincing ... but something must be wrong, because the area of Innumeratus’s ‘square’ must be less than that of the original ‘square’. In fact, neither shape is a perfect square. The original one bulges slightly outwards in the middle; the second bulges slightly inwards. For example, the two different sized triangles have horizontal and vertical sides in the ratios 8:3 and 5:2 respectively. If the figure were square, these ratios would be equal. But they are 2.67 and 2.5, which are different.
Pieces of Five
The bosun placed one coin on the table, and then put two on top of it so that they touched at its centre. To do this, hold them in place while you fit the other two coins almost on edge, leaning together to touch at the top. Again, all coins touch, so in particular they are equidistant.
Place the first three coins as on the left, then add the other two.
The Curious Incident of the Dog
The next number in the sequence is 46.
Holmes’s point is: don’t look at what’s there, look at what’s missing. The missing numbers are:
3 5 6 9 10 12 13 15 18 20 21 23 24 25 27
30 31 32 33 34 35 36 37 38 39 40 42 43
These are the multiples of 3, the multiples of 5, anything containing a digit 3, and anything containing a digit 5. The next number in the sequence is therefore 46 (because 45 is a multiple of 5).
Mathematics Made Difficult
Lagrange’s interpolation formula states that the polynomial
satisfies P(xj) = yj for j = 1, ... , n. Remember: Linderholm’s book is based on the premise that mathematics should be made as
complicated as possible to enhance the prestige of the mathematician. Actually, the basic idea here is simple. In less compact notation, the formula becomes
When x = xj, all terms except the jth vanish because of the factor (x - xj). The jth term doesn’t have that factor, and it is an apparently complicated fraction times yj. However, the numerator and denominator of the fraction are identical, so the fraction is 1. And 1 times yj is yj. Cunning!
For example, to justify the sequence 1, 2, 3, 4, 5, 19, we take x1 = 1, x2 = 2, x3 = 3, x4 = 4, x5 = 5, x6 = 6 and y1 = 1, y2 = 2, y3 = 3, y4 = 4, y5 = 5, y6 = 19. Then a calculation gives
and P(1) = 1, P(2) = 2, P(3) = 3, P(4) = 4, P(5) = 5, P(6) = 19.
Edward Waring first published the formula in 1779. Euler rediscovered it in 1783, and Lagrange discovered it again in 1795. So it is named after the third person who found it, which is fairly typical when it comes to naming mathematical ideas after people.
A Four Colour Theorem
These 11 circles require four colours.
Eleven circles are needed. Here’s an arrangement that requires four colours. Suppose for a contradiction that it can be 3-coloured. Colour the top middle circle with colour A, and the two adjacent ones with colours B and C. Then the next circle to the left must have colour A, the one below that colour B, and the left-hand circle of the two lowest ones must have colour A. The colours round the right-hand side of the figure must be similar: here either X = B and Y = C, or X = C and Y = B. Either way, the right-hand circle of the two lowest ones must have colour A. But now two adjacent circles have the same colour, namely A - contradiction.
It can be proved that with ten or fewer circles, at most three colours are needed.
Serpent of Perpetual Darkness
The Earth goes round the Sun exactly once in a year, so it returns to the same point in its orbit on any given date (subject to a bit of drift because the exact period is not an integer number of days, whence our use of leap years). In particular, every 13 April it returns to a position at which the orbit of Apophis crosses that of Earth, a necessary condition for a collision.
Apophis has a period of 323 days, and the distance between Apophis and the Sun ranges from 0.7 astronomical units to 1.1 astronomical units, where an astronomical unit is the average distance between the Sun and the Earth. So sometimes it is inside Earth’s orbit, sometimes outside. If Apophis and the Earth orbited in the same plane, their orbits would cross at two points. However, it’s not quite that simple. The orbits are inclined at a small angle, and Apophis is light enough to be affected significantly by the gravitational pull of the other planets, causing changes to its orbit. So although the orbits don’t necessarily cross, the orbit of Apophis - though not necessarily the asteroid itself - comes close to that of the Earth in two places, and the Earth reaches those positions on two specific dates. The one that counts, for the near future, is where the Earth is every 13 April. Whether there’s a collision depends on precisely where Apophis is in its orbit on that date, and a lot of high-precision observations are needed to determine that. So the date is easy, but the year is difficult.
Orbit of Apophis relative to that of the Earth.
Actually, there was more to it. There always is. Calculations showed that if Apophis happens to pass through a particular region of space, about 600 metres across, during a near-miss in 2029, then it is guaranteed to return to almost the same spot, hitting the Earth in 2036. Fortunately, the latest observations indicate that the chance of such a collision is at most 1 in 45,000. See neo.jpl.nasa.gov/news/news146.htmlscience.nasa.gov/headlines/y2005/13may_2004mn4.htm
Both of these refer to Apophis by its provisional designation 2004 MN4.
What Are the Odds?
No, she’s wrong, the probability is . And it’s very naughty of her to try to swindle poor Innumeratus like that.
Whichever card he picks first, the remaining cards include two of the opposite colour but only one of the same colour. So the probability of picking a card of the opposite colour is . Since this holds whichever card he picks first, the probability that his two cards are different is .
Here’s another way to see it. There are 6 distinct pairs of cards. Of these, precisely 2 (♠ ♣ and ♥ ♦) have the same colours, and 4 have different colours. So the probability of getting one of these four is .
The Shortest Mathematical Joke Ever
In analysis, ε is always taken to be a small positive number, to the point of cliché. So this joke is a more intellectual variant on the mechanics question that began ‘An elephant whose mass can be neglected ...’
Name the Cards
The cards were the King of Spades, Queen of Spades and Queen of Hearts. The first has to be a Spade and the third a Queen, but the precise sequence is not determined.
The first two statements tell us that the cards must either be KQQ or QKQ.
The last two statements tell us that the cards must either be ♠ ♠ ♥ or ♠ ♥ ♠.
Combining these we find four possible arrangements:
The fourth of these contains the same card twice, so it is ruled out. The other three all use the same three cards, in various orders.
This puzzle was invented by Gerald Kaufman.
Nice Little Earner
Surprisingly, Smith earned more - even though £1,600 per year is greater than Smith’s accumulated £500 + £1,000 over a year. To see why, tabulate their earnings for each six-month period:
Smith Jones
Year 1 first half £5,000 £5,000
Year 1 second half £5,500 £5,000
Year 2 first half £6,000 £5,800
Year 2 second half £6,500 £5,800
Year 3 first half £7,000 £6,600
Year 3 second half £7,500 £6,600
Note that Jones’s £1,600 splits into two amounts of £800 for each half-year, so his half-year figures increase by £800 every year. Smith’s half-year figures increase by £500 every half-year. Despite that, Smith is ahead in every period after the first, and gets ever further ahead as time passes. In fact, at the end of year n, Smith has earned a total of 10,000n + 500n(2n - 1) pounds, while Jones has earned a total of 10,000n + 800n(n - 1) pounds. So Smith - Jones = 200n2 + 300n, which is positive and grows with n.
A Puzzle for Leonardo
Emperor Frederick II was seeking a rational number x such that x, x - 5 and x + 5 are all perfect squares. The simplest solution is
for which
Leonardo explained his solution in 1225 in his Book of Squares. In modern notation, he found a general solution
Here the role of x is played by the number (m2 + n2), and we want mn(m2 - n2) = 5. Choosing m = 5, n = 4, we get x = 3 and mn(m2 - n2) = 180. This may not seem much help, but 180 = 5 × 62. Dividing x by 6 yields the answer.
It’s About Time
Crossnumber solution.
Do I Avoid Kangaroos?
I avoid kangaroos.
Write the conditions symbolically, as on page 275. Let
A = avoided by me
C = cat
D = detested by me
E = eats meat
H = in this house
K = kangaroos
L = loves to gaze at the moon
M = kills mice
P = prowls by night
S = suitable for pets
T = takes to me
Then with ⇒ meaning ‘implies’ and ¬ meaning ‘not’, the statements (in order) become
H ⇒ C, L ⇒ S, D ⇒ A, E ⇒ P, C ⇒ M
T ⇒ H, K ⇒ ¬ S, M ⇒ E, ¬ T ⇒ D, P ⇒ L
Now we appeal to the laws of logic that I mentioned on page XXX:
X ⇒ Y is the same as ¬ Y ⇒ ¬ X
If X ⇒ Y ⇒ Z, then X ⇒ Z
Using these laws, we can rewrite these conditions as
¬ A ⇒ ¬ D ⇒ T ⇒ H⇒ C ⇒ M ⇒ E ⇒ P ⇒ L ⇒ S ⇒ ¬ K
so that ¬ A ⇒ ¬ K, or equivalently, K ⇒ A.
Therefore I avoid kangaroos.
The Klein Bottle
&
nbsp; To cut a Klein bottle into two Möbius bands, slice it lengthwise, cutting through the ‘handle’ of the bottle and the body along the plane of mirror symmetry. A little thought shows that the two pieces are Möbius bands.
Cutting a Klein bottle into two Möbius bands.
Accounting the Digits
This is the only number that works.
As Long as I Gaze on Laplacian Sunrise
If Laplace’s figures are correct - which is highly debatable - the probability that the Sun will always rise is zero.
The probability of the Sun rising on day n is (n - 1)/n. So:
• The probability of the Sun rising on day 2 is
• The probability of the Sun rising on day 3 is
• The probability of the Sun rising on day 4 is and so on. Therefore
• The probability of the Sun rising on days 2 and 3 is
• The probability of the Sun rising on days 2, 3 and 4 is
Ian Stewart Page 25