Solution of the Week #368 - Eleventh Root

While the question says that it must be solved in your head, I’m obviously having to present it here on paper, but hope you’ll agree that the individual steps can be done mentally.

First of all we want to know how many digits in N. If N was 10, N^11 would be 1 followed by eleven zeroes. If N was 100, N^11 would be 1 followed by twenty-two zeroes. Since the N^11 in the question has 19 digits, N must be between 10 and 100, and so therefore has two digits.

Let’s find the last digit first. The final digit of N^11 is 3. Clearly the digit we seek cannot be even. We can try out each of the odd digits. To do this we can repeatedly multiply by a digit and discard all but the final digit, until we get back to where we started. Powers of 5 always end in 5. Powers of 1 always end in 1. Powers of 9 alternate between 9 and 1. So we are left with 3 or 7. 3 follows the repeating pattern 3 - 9 - 7 - 1, and 7 follows the repeating pattern 7 - 9 - 3 - 1, so both could end in a 3. But we are looking for an eleventh power, and since both these sequences are four in length, it is the third in the repeating sequence that we are interested in. Therefore the final digit of N is 7.

Next, since 11 is prime, we can make use of Fermat’s Little Theorem to find out the divisibility of N with respect to 11. The test for divisibility by 11 is very easy: add and subtract alternate digits, and if the result is a multiple of 11, then the original number was too. Determining modulo 11 is similar but now we must make sure that the final digit is added, so in other words all odd-positioned digits need to be added and all even-positioned digits subtracted.

Working through the given number in order:

2-4+7-2+1-5+9-2+1-5+0-8+4-0+1-2+3-0+3 = 3

Since N^11(mod 11) = N(mod 11), we know that N must be three more than a multiple of 11. Since we already know the final digit is 7, we can tell that the first digit is 4.

Therefore N = 47.

Solution of the Week #367 - Come Close But Don't Touch

It turns out the best approach is a kind of ‘greedy’ algorithm, where each unit fraction is the most it can be, given how much is left of the 1 we are trying to almost fill. This is easily calculated: if there is 1/n remaining, the next biggest unit fraction will be 1/(n+1).

So the first unit fraction will be 1/2, leaving 1/2.

Therefore the second unit fraction is 1/3, leaving 1/6.

Next is 1/7, leaving 1/42.

Next is 1/43, leaving 1/1806.

Finally we add 1/1807, leaving us just 1/3263442 short of the 1.

So our final sum comes to 3263441/3263442 = 0.9999996936…

Solution of the Week #366 - Wordsalad

ANGST, ATONE, OASIS, OLIVE, RADIO, ROBOT.

Solution of the Week #365 - Twenty Coloured Balls

Instead of thinking about probability, think in terms of how many different pairs are possible. Since each pair is equally likely, counting the pairs and comparing them is all we need to do.

There are 20 balls altogether, so there are 20 options for the first ball. There are 19 options for the second ball, but since a pair is the same pair if they are drawn in reverse order, we can halve this product. So the total number of possible pairs is 20 x 19 / 2 = 190.

We can perform the exact same calculations for each set of same-coloured balls within the bag. For instance if there were 15 red balls, then there are 15 x 14 / 2 = 105 pairs that would be made up of two red balls. This is already more than half of 190, so there cannot be 15 balls of a particular colour.

Instead let’s try 14 red balls. This gives us 14 x 13 / 2 = 91 red-red pairs. This is close to the 190 / 2 = 95 pairs we need, so we are on the right track.

If there were 3 balls of a second colour (say, blue) then there would be 3 x 2 / 2 = 3 possible blue-blue pairs. Add these to the 91 red-red pairs and we are almost there.

Add in 2 balls of a third colour (say, white) then there is one white-white pair. This makes up all of the 95 pairs we needed, but we have only used 19 balls, so we need a single ball of a fourth colour (say, green) to make it up to 20.

14 RED, 3 BLUE, 2 WHITE, 1 GREEN.

If you play about with other possibilities you will find that this arrangement, (14,3,2,1), is the only way of exactly achieving the aim of exactly 50% chance of a matching pair.

Solution of the Week #364 - Points on a Circle

We can start off by saying that c>=b>=a, and worry about reflections and rotations later.

First, we’ll deal with situations where some of the lengths are equal. Clearly they can’t all be equal, as the point would be the centre of the circle, not on it. If a=b, then c would be a diameter of the circle, which is 49*(2/rt(3)) ~ 56.5, not an integer. If b=c, then a=0 and we have a vertex of the original triangle. There are three vertices, so already three points which satisfy the question.

We can now make use of a couple of facts about cyclic quadrilaterals: by Ptolemy’s theorem 49c = 49a+49b, so therefore a+b is always equal to c, which is convenient since if a and b are both integers, c will be too. Because opposite angles of a cyclic quadrilateral sum to 180, the angle between a and b is 120 degrees.

Now using the fact that cos(120) = -0.5, we can use the cosine rule to find the following equation:

49^2 = a^2 + b^2 + ab

If we add ab to both sides, then factorise the right-hand side we get:

49^2 + ab = (a+b)^2

a+b=c must be between 49 and the diameter ~56.5, so there are only 7 possible integer values to try, and each will give a different value for ab

(a+b,ab) = (50,99),(51,200),(52,303),(53,408),(54,515),(55,624),(56,735)

Only the last two of these have solutions in the integers:

(a,b,c) = (16,39,55) and (21,35,56).

In our diagram a is the distance from the top vertex, b from the left vertex and c from the right vertex, but these vertices can be permuted in 3! = 6 ways, so these two numerical solutions represent 12 points on the circle. Along with the three vertex points themselves there are therefore 15 points on the circle that are an integer distance from all three vertices of the triangle.

Solution of the Week #363 - Zero to Pi(ish)

The easiest thing is to do the whole procedure in reverse. Action a then become x-1 (action b is unchanged).

So starting with a rational number, to get to 0 in the fewest steps, whenever your number is greater than or equal to 1, perform action a and whenever your number is less than 1, perform action b.

In other words, each time the numerator is less than the denominator, flip the fraction upside down, and in doing so you will successively reduce the denominator until it is 1 and you can then keep subtracting until you get to zero.

From 355/113 to 0 will take a total of 28 actions:

355/113 = 3 16/113

a, 3 times

16/113

b

113/16 = 7 1/16

a, 7 times

1/16

b

16

a, 16 times

0

Reversing the process to go from 0 to 355/133 will clearly also take 28 actions: 16a,b,7a,b,3a.

Solution of the Week #362 - Missing Integers

2^a + 2^10 + 2^13 = b^2

Looking first just at 2^10 + 2^13, since 10 is less than 13 we can take out a factor of 2^10 to get:

2^10(1 + 2^3)

2^10 is obviously (2^5)^2 and (1 + 2^3) is 9, also a square number, therefore, you can replace 2^10 + 2^13 by (3*2^5)^2, or 96^2

2^a + 96^2 = b^2

Moving this to the right-hand side we have:

2^a = b^2 - 96^2

Factoring the difference of two squares gives:

2^a = (b+96)(b-96)

This means that both of the factors on the right-hand side are powers of 2, let’s call them 2^c = b+96 and 2^d = b-96.

If we take their difference to eliminate b we get:

2^c - 2^d = 192

Since d is smaller than c, we can take out a factor of 2^d:

2^d(2^(c-d)-1)=192

which is a product of a power of 2 and an odd number, which must therefore be 2^6 and 3.

d is therefore 6, and c is 8. a, as the sum of c and d, is therefore 14, and b is 2^6 + 96 = 160.

So the completed equation is

2^14 + 2^10 + 2^13 = 160^2

Solution of the Week #361 - Round and Round

We can get an approximate value for x by ignoring the rounding steps temporarily.

5*4*3*x^3 ~ 273

x^3 ~ 4.55

x ~ 1.657…

We know that the round(4x(round(3x)) part has been rounded to an integer, so to try to discover exactly what integer it is we can plug our approximate value of x to the outside of it:

5(1.657)(integer) ~ 273

integer ~ 273/(5(1.657)) ~ 32.95

This is close to an integer, 33, but we need to test it by plugging it back in:

Assume the value of round(4x(round(3x)) is indeed 33:

5x*33=273

x = 273/(5*33) = 273/165 = 91/55

Plugging that value into round(4x(round(3x)) indeed gives 33, so this is the correct answer.

x = 91/55 = 1.6545454..

Since this is a strictly increasing function (except where 3x rounds to zero), once we have found an answer, we can be sure it is the only answer.

General solution to 9-point circle problems

My two previous puzzles have been based on the nine-point circle of a triangle. They both stemmed from a series of equations I worked out for the angles between the nine-point centre and each of the nine points. The following is based on an acute triangle, as has been shown you can swap a vertex with the orthocentre to get to that situation. Every third point on the circle is the midpoint between a vertex and the orthocentre, and the order you encounter midpoints / altitude feet along each side depends only on the order of the angle sizes.

As you can see, at least four of those angles will be the same. In puzzle 359, the apex angle was the lowest angle ‘c’ and ‘a’ and ‘b’ were equal. In puzzle 360, it was ‘b’ and ‘c’ that were equal and ‘a’ was at the apex.

Solution of the Week #360 - Isosceles Nine-Point Circle part 2

The angle at the apex will be 72 degrees.

The two longer lengths are two sides of a regular pentagon, and the six shorter sides are sides of a regular decagon.

The ration of long length to short length is sin(36)/sin(18) = sqrt(1/2(5+sqrt5)) ~ 1.902…

Solution of the Week #359 - Isolsceles Nine-Point Triangle

The angle at the apex will be 45 degrees.

I alluded to the existence of a second possible solution. An interesting fact is that if you consider the orthocentre of any triangle along with the three vertices, you can in fact choose any three of those four points to form a triangle, and the resulting nine point circle will be the same, with the nine points in all the same places (although midpoints might become ‘halfway to orthocentre’ points and vice versa). So this means there are four possible triangles with the exact same set of eight points as above. However, since we are specifically asked for an isosceles triangle, there is only one additional triangle which fits the bill, ie the triangle formed by the two lower points of the above triangle, along with what was the orthocentre. This has an apex angle of 135 degrees.

Solution of the Week #358 - Power Dates

The only date that’s a power higher than 5th is Thursday 13th October 2072, which becomes 131072, which is 2^17.

Solution of the Week #357 - First Date

Of the days of the week, Friday is first in alphabetical order, and amongst the months April is first. When the date is written out in full, ‘eighteenth’ is first alphabetically.

The final day, date and month alphabetically will respectively be Wednesday, twenty-third and September, so the answer is Wednesday 23rd September.

Solution of the Week #356 - Counter Game 2

This can be solved using exactly the same method as the first Counter Game puzzle. For all negative numbers, the number of ways is 0, and for 0 it is 1, and each subsequent number n, the number of ways is the sum of the number of ways of n-2 and n-7.

0:1

1:0

2:1

3:0

4:1

5:0

6:1

7:1

8:1

9:2

10:1

11:3

12:1

13:4

14:2

15:5

16:4

17:6

18:7

19:7

20:11

21:9

22:16

23:13

24:22

25:20

26:29

27:31

28:38

29:47

30:51

31:69

32:71

For 25 counters the number of ways is less than for 24. By coincidence n=25 is also the last time that the number of ways is less than n. To know for sure the sequence doesn’t reverse again after this point, it is only necessary to find 7 in a row that are increasing, hence me continuing the sequence to n=32.

Solution of the Week #355 - Human Logic

This puzzle will necessarily have several possible answers; here is one I came up with. To reach 9 words with only 10 different letters, each word in the list should only introduce one new letter, with the exception of the first word, although even that needs to be limited to just two different letters.

MAMMA MA

MAGMA G

LLAMA L

HALAL H

LAUGH U

GHOUL O

AMIGO I

HUMAN N

LOGIC C

Solution of the Week #354 - Counter Game

10609 different ways. This is 103 squared.

Trying to go ahead and systematically list all of the ways is going to be a massive task, but luckily there is a quicker way.

The first go can either be 1, 2 or 3. If it was 1 then there would be 15 counters left; if it was 2 there would be 14 counters left; if it was 3, 13 left. All quite obvious so far. So if we only knew how many ways there were of combining to make 13, 14, and 15, we could just add them together to find the total number of ways for 16. We can do this safely knowing there is no overlap but that they cover all cases, as they all start with different numbers. All very nice, but the trouble is we don’t know how many ways there are of making 13, 14 or 15. Luckily our rationale will work in general, and for any number of board spaces, we only have to add up the previous three to find the answer. This holds until the number of counters is less than four, but luckily those numbers are more manageable.

1:1 way

2:2 ways (1+1 or 2)

3:4 ways (1+1+1, 1+2, 2+1, 3)

Building on this, knowing that each subsequent number of ways is simply the sum of the previous three, we can soon find the answer:

4:7

5:13

6:24

7:44

8:81

9:149

10:274

11:504

12:927

13:1705

14:3136

15:5768

16:10609

There is even a clever way to avoid working out the first few, that works in general. For each negative number there are 0 ways, and for zero there is 1 way, then every subsequent number of ways is the sum of the previous n. So for instance if n=6, ie. if you had a six-sided dice and wanted to know how many different ways of reach exactly 10 with multiple throws:

0:1

1:1

2:2

3:4

4:8

5:16

6:32

7:63

8:125

9:248

10:492

As an extra bonus question, does this sequence, the one based on a six-sided dice, ever yield a prime number of ways (apart for the 2 ways of getting 2)?

Solution of the Week #353 - Odd Shape

The squares have area 4, 9 and 31.25 respectively, and the triangles have areas of 2.25, 1 and 2.5. The total area is 50 square units.

Solution of the Week #351 - Secret Message

Within each block, countries can be read diagonally downwards, jumping back to the top after the bottom row of the block. The message reads: YOU ARE OK.