Solution of the Week #409 - Classes

If we say that class B has ‘x’ girls, and class C has ‘y’ boys, then class A has 2x girls and 4y boys. Also, we are told that class C has 2 more girls than class B has boys, but since the classes are equal that means class B has 2 more girls than class C has boys. So x = y+2. Class A has 2(y+2) girls and 4y boys, so 6y+2 students in total. Again since all classes are equally sized, class B has y+2 girls and therefore 5y+2 boys, and class C has y boys and therefore 5y+4 girls. In total there are 8y+10 girls and 10y+2 boys. Since we are told these numbers too should be equal, y=4.

Therefore class A has 12 girls and 16 boys, class B has 6 girls and 22 boys, and class C has 24 girls and 4 boys. There are 84 students in total.

 

Solution of the Week #407 - Strange Area Dissection

It is possible for this shape to tile the infinite plane, if copies of the shape are rotated 45, 135, 225 and 315 degrees from the original. You can then overlay a unit square grid which neatly lines up with the shape as below:

Translating this back to the original shape, this corresponds to drawing lines from the centre of the semicircle to each of the 90 degree corners, dissecting the shape into two isosceles triangles and a dart shape. Not only is this a dissection in only three pieces, but it can be hinged as well, a la Henry Dudeney.

Solution of the Week #405 - Number Hunt

All prime numbers except for 2 and 3 are either of the form 6n+1 or 6n-1.

A number of the form 6n+1 can be expressed as 9+25+3(2n-11). The first two numbers are obviously composite. The last is composite whenever 2n-11 is at least 3, so when n is 7 or more. So 43, 49, 55, 61, etc. The highest prime number in this class below 43 is 37.

A number of the form 6n-1 can be expressed as 9+35+3(2n-15). Again the first two numbers are obviously composite. The last is composite whenever 2n-15 is at least 3, so when n is 9 or more. So 53, 59, 65, 71, etc. The highest prime number in this class below 53 is 47.

Therefore the answer is 47.

For completeness, the full list of prime numbers that cannot be expressed as the sum of three positive composite odd numbers is:

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41 and 47.

Solution of the Week #399 - What's Wrong

That OAB and BDC are equal presupposes that AC and BD are parallel. This is an unwarranted assumption.

Instead we should work out what the angle AOB is. If OAB is indeed 32 degrees, then to complete the 180 degrees of the triangle OAB, angle AOB should be 58 degrees.

Alternatively, if we assume CDB is indeed 32 degrees, then from the central angle theorem, AOB is equal to twice 32: 64 degrees.

So both marked angle being 32 degrees leads to a contradiction.

If OAB is 32 degrees, BDC is 29 degrees, whereas if BDC is 32 degrees, OAB is 26 degrees. The only time they can be equal is if they are both equal to 30 degrees.

Solution of the Week #398 - Find the Radius

If we draw a line from the origin, directly through the centre of the circle and to the point where the arc and the circle meet, that line is clearly equal to R, but it is also equal to (sqrt(2)+1)*r, where r is the radius of the circle. So we have R = (sqrt(2)+1)*r, and if we square both sides (you’ll see why in just a moment) we get:

R^2 = (2*sqrt(2)+3)*r^2

We can also form a right-angle triangle using the origin and the line of length 1, to get:

R^2 = (2r)^2 + 1 = 4r^2 + 1

So we have two alternative equations for R^2, which we can equate:

(2*sqrt(2)+3)*r^2 = 4r^2 + 1

Making r^2 the subject gives:

r^2 = 1/(2*sqrt(2)–1)

Which we can make the denominator rational by multiplying top and bottom by 2*sqrt(2)+1:

r^2 = (2*sqrt(2)+1)/7

We can find R^2 by using either of the relations we worked out previously:

R^2 = 4(2*sqrt(2)+1)/7 + 1

R^2 = (11+8*sqrt(2))/7

Finally R= sqrt((11+8*sqrt(2))/7) ~ 1.7854

 

Now for the 3-point solution, let R^2 = x

x = (11+8*sqrt(2))/7

but we would like it to look like the quadratic formula (-b +/- sqrt(b^2-4ac))/2a , so the first thing we need is an even number in the denominator, and we also need to bring the 8 under the square root:

x = (22+sqrt(512))/14

So, a = 7, b = -22, and b^2-4ac = 512, 484 - 28c = 512, therefore c = -1

So the polynomial in x is:

7x^2 – 22x - 1

And back in terms of R is:

7R^4 - 22R^2 - 1

This polynomial has four roots, however two are complex, and one is negative, and so the only positive real root is the value for R that we previously found.

 

 

Solution of the Week #397 - Starting Squares

16974441796

1 = 1^2

16 = 4^2

169 = 13^2

169744 = 412^2

16974441796 = 130286^2

 

1,4 or 9 would satisfy the first part

16 or 49 would satisfy the next part

169 is the only number that satisfies the next part

To proceed we need a square number that lies between 169000 and 169999. By taking the square root of both those numbers, we find the only whole number between them is 412, therefore 412^2 = 169744 is the only number which satisfies the next part.

Similarly, we take the square root of 16974400000 and 16974499999 and find only one whole number between them: 130286. Therefore the answer is 130286^2 = 16974441796.

Solution of the Week #395 - Vaguely Defined Pentagon

The first thing to determine is where on the shape are the three right angles. There are only two possible cases: the two non-right angles are adjacent, or they are not.

Let’s consider the case where they are adjacent, as above. Clearly t>r and s>p. The only way this is possible is if q is the second shortest length. q is the hypotenuse of a triangle whose other sides are (s-p) and (t-r). One of these will be equal to 2 and the other 3, and so q equals sqrt(13). This is not a whole number, and so therefore this case is eliminated.

In this second case, the dashed line is simultaneously the hypotenuse of q and r, and of t and (s-p). Checking through the cases where t is the shortest, second shortest, or third shortest side, there are no solutions. I’ll demonstrate this by assuming that it is the third shortest. As I’ve shown s>p, we’ll consider q=n, p=n+1, t=n+2, s=n+3, r=n+4.

The square of the dashed line = n^2 + (n+4)^2 = (n+2)^2 + 2^2

Expanding each squared bracket gives:

2n^2+8n+16 = n^2+4n+8

Clearly the left hand side is always greater than the right, for any positive whole number n.

Next let’s consider when t is the longest length:

p=n, q=n+1, r=n+2, s=n+3, t=n+4.

The square of the dashed line = (n+1)^2 + (n+2)^2 = 3^2 + (n+4)^2

n^2+2n+1 + n^2+4n+4 = 9 + n^2+8n+16

n^2 – 2n – 20 = 0

Using the quadratic formula, and eliminating the case when n is negative, we get n = 1+sqrt(21). Clearly not a whole number.

Finally let’s look at the last remaining case, where t is the second longest side. r=n, q=n+1, p=n+2, t=n+3, s=n+4.

The square of the dashed line = n^2 + (n+1)^2 = (n+3)^2 +2^2

2n^2+2n+1 = n^2+6n+13

n^2 – 4n – 12 = 0

n = (4 + sqrt(64))/2 = 6

So we now know that the side lengths are 6,7,8,9,10 starting at r and going anti-clockwise. Finally, to find the area, we work out the area of the trapezium to the left of the dashed line and the triangle to the right: 81+21 = 102. And that’s the answer!

 

Solution of the Week #394 - Coin Paradox

Statements 1, 2 and 3 are true.

Statement 1: HTH and HTT will each happen 1/8 of the time.

Statement 2: at some point HT will come up on consecutive throws. The next throw will be H or T with equal probability and so HTH and HTT will have an equal chance of coming up.

Statement 3: the average number of throws to see HT is 4, and as we have seen, the game will finish on the next throw regardless of what it is, so the average number of flips ending in either HTH or HTT is 5.

Statement 4: this seems like it should be true, but it isn’t. On average it will take me 10 throws to see HTH but my friend only 8 throws to see HTT. This is because if my friend reaches HT, she will either finish on the next throw, or have an H to start off a new attempt at HTT. If I’m on HT I will either finish on the next throw or be back at square one.

 

Interestingly, if we make the target sequences differ earlier, for instance HHH vs HTT, then even statements 2 and 3 will be false. HTT would win 3 times out of 5, and the average number of flips when HHH wins would be 5 2/5, whereas the average number of flips when HTT wins would be 5 11/15.

 

Solution of the Week #393 - Two Point Oh Eight

Let’s begin by seeing what happens with four digits, five digits, etc.

With four digits there are two numbers that work: 5252 and 5772.

Since in each case abcd/dcba is equivalent to 52/25, abcd must be a multiple of 52 and dcba must be a multiple of 25. The same multiple in fact.

Let’s take a look at exactly what that multiple is for the first few:

52/25 = 1

572/275 = 11

5252/2525 = 101

5772/2775 = 111

52052/25025 = 1001

57772/27775 = 1111

You might notice that the length of each of these factors is always one less than the number of digits of the final number, because multiplying by 52 or 25 increases the digits by 1. You might also notice that each number is palindromic, because the denominator needs to be the reverse of the numerator. And that it only uses digits 1 and 0, because if you had any digits greater than 1, there would be carrying, which would disrupt the digits of the numerator and the denominator in different ways.

For the 25-digit numbers of the question, we need to find all palindromic 24-digit numbers that are made up of 1s and 0s.

The first digit needs to be a 1 (otherwise the number would not be 25 digits long). The next 11 digits can be either 0 or 1, and the remaining 12 digits are defined by the fact that it needs to be a palindrome.

So we have 11 binary decisions to make. 2^11 = 2048. Therefore there are 2048 25-digit numbers which, when divided by their reversal, equal 2.08. You’ll forgive me for not listing them all, but I’ll pick one at random by flipping a coin:

100010110010010011010001:

5200525720520520572520052/2500252750250250275250025 = 2.08

 

 

Solution of the Week #392 - Shaded Areas

If we let the dimensions of the rectangle be 2x and 2/x, then whatever the value of x, its area will be 4.

This means that the radii of the three semicircles will be:

x, 1/x and (x+1/x)

The area of a semicircle is (pi*r^2)/2, and we want the area of the larger semicircle, less the area of the smaller two. For simplicity we can take out the pi/2:

Area = pi/2 * ((x+1/x)^2 – x^2 – 1/x^2)

Area = pi/2 * (x^2 + 2(x/x) + 1/x^2 – x^2 – 1/x^2)

x gets cancelled out

Area = pi/2 * (2)

Area = pi

And that’s the answer!

Solution of the Week #391 - Five Towns 2

The original puzzle was solved using Pythagorean triples. It’s impossible to find a Pythagorean triple where none of the numbers is divisible by 3.

Instead we can use the cosine rule, which is a kind of generalised version of Pythagoras theorem.

 

a^2 + b^2 - 2ab*cosC = c^2, where C is the angle opposite the side c. (When C is 90, cosC is zero and that whole term disappears).

 

Clearly if a b and c are integers, 2ab*cosC needs to be too. cosC can be anything from 1 to -1. We don’t care if C is rational (in whatever units: degrees, radians, etc), only that cosC is rational.

We need two adjacent central angles to add to 180 degrees. It turns out, conveniently for our purposes, that if C and C’ add to 180 degrees, then cosC and cosC’ add to zero.

 

So now we just need to find a convenient rational value for cosC, for example 1/13, then find a triple a,b,c that satisfies

a^2 + b^2 - 2ab/13 = c^2

none of which is a multiple of 3, for example 13,20,23

and other triples that satisfy

a^2 + b^2 + 2ab/13 = c’^2

with again, no multiples of 3, for example 10,13,17 and 13,40,43

Making a double size copy of 13,20,23 and also doubling 10,13,17 gives us four triangles of alternating central angles, which fit together as shown:

I thank Philip Morris Jones for this solution.

There is one more solution with a total mileage of less than 400 miles. It is based on a value of cosC of 1/5, and triples of 4,5,7 and 7,10,11, with roads leading from the centre with lengths of 50,40,28 and 35, and around the outside of 70,44,49 and 55, giving a total mileage of 371.