Solution of the Week #575 - 166669

We only need to look at prime number, since if a number isn’t divisible by p, it won’t be divisible by any multiple of p.

We’ll start at the bottom and work our way up. Since the number always ends with 9, it will never be divisible by 2.

For divisibility by 3, the sum of the digits would need to be divisible by 3, but since the digits 6 and 9 are divisible by 3, the leading 1 means the number as a whole will never be.

Since the number ends with 9 it will never be divisible by 5.

Next let’s look at 7. If we multiply a number of the form 16..9 by 3 we will always get a number of the form 50..7. if we then subtract 7 we will have a number that is clearly just the product of 2s and 5s. Since multiplying by 3 and subtracting 7 would have preserved any divisibility by 7, this proves that 16..9 is never a multiple of 7.

For divisibility by 11, the sums of alternate digits need to either be equal, or differ by a multiple of 11, for example 1342 is divisible by 11 because the first and third digits add to the same number as the second and fourth digits.

Let’s consider when we have an odd number of digits. Then 1+k+9 would need to equal (or differ by 11n) 6+k. k in each case is some indeterminate number of 6s, but would cancel out in this case. 10 – 6 is 4, not a multiple of 11. But if the number of digits is even we have 1+k and k+9. The difference here is 8, again not a multiple of 11. So 166…9 can never be a multiple of 11.

Finally 169 is 13 x 13, and so 13 is the smallest number that divides a number from this sequence.

Solution of the Week #573 - 127 Cubes

The simplest way is to start by dissecting it by 4x4x4 = 64. If you then dissected one of these 64 cubes into 2x2x2=8 cubes you have increased the number of cubes by 7. You can repeatedly do this, either to more of the original 64 cubes, or even some of the new smaller cubes. Either way, dissecting 9 cubes into 8 will increased the number of cubes from 64 to 127.

In fact you can dissect a cube into almost any number of smaller cubes. There is just a short list of small numbers that cannot be achieved, the highest being 47.

Solution of the Week #572 - Quadrilteral

The most efficient way of enclosing the circles is first to arrange them so that each circle is tangent to the other two, draw an isosceles triangle around them, and then draw a fourth side, tangent to larger circle, but parallel to the shorter side of the isosceles triangle.

(The other candidate shape is placing the circles in a row and building a longer slimmer trapezoid around them, but for a quick sanity check, that shape would have to be greater in area than 196^2 + 144^2 + 144^2, which is 79888, and as we shall see the shape below is smaller).

 Using coordinate geometry I ascertained the dimensions of this trapezoid. I won’t bore you with the messy details, especially as there’s doubtless a far more elegant way to do it. In any case the base measures 336, the top 147, the sloping sides 337.5, the height 324 and therefore the area is 78246 square units.

Solution of the Week #571 - Crossing Triangles

Take the right-way-up large equilateral triangle. If we work out what the area of it is we can just subtract 3, 12 and 27 to find the hexagon area.

The relationship between area A and side length s of an equilateral triangle is A = sqrt(3)/4 * s^2

The reverse of this is that s = sqrt(4A/sqrt(3))

So the side length of the 12, 48 and 27 triangles are respectively approximately 5.264, 10.529 and 7.896, summing to about 23.689.

Then the area of the right-way-up large triangle comes out as exactly 243.

The hexagon area is therefore exactly 201

Solution of the Week #570 - Cube

There must be a better way to solve this but this is what I did:

 If we call the side length x, and the coordinates of the point in space relative to the middle vertex, (a,b,c) we can form four equations:

 a^2 + b^2 + c^2 = 583^2

(a+x)^2 + b^2 + c^2 = 637^2

a^2 + (b+x)^2 + c^2 = 713^2

a^2 + b^2 + (c+x)^2 = 727^2

Subtracting the first equation from each of the other three in turn gives:

2ax + x^2 = 65880

2bx + x^2 = 168480

2cx + x^2 = 188640

Rearrange each to isolate a,b,c

a = (65880 - x^2)/2x = 32940/x - x/2

b = (168480 - x^2)/2x = 84240/x - x/2

c = (188640 - x^2)/2x = 94320/x - x/2

Let’s square each of those expressions

a^2 = 32940^2/x^2 – 32940 + x^2/4

b^2 = 84240^2/x^2 – 84240 + x^2/4

c^2 = 94320^2/x^2 – 94320 + x^2/4

Adding them together and equating the sum to 583^2:

(32940^2+84240^2+94320^2)/x^2 – 211500 + 3(x^2)/4 = 583^2

Multiply every term by 4x^2 to eliminate the x^2 term in the denominator:

Solve the quadratic in x^2 to x^2 = 32400 or 2108356/3, so x is either 180 or 838.322929…, however in this second case the point would be inside the cube.

The cube’s side length is therefore 180

Solution of the Week #568 - Dice

Each sequence will continue until I roll either a 1 or a 6, so will end with a probability of 1/3, therefore making the average length 3. The fact that the sequences that end in a 1 instead of a 6 are discarded is irrelevant for the average sequence length.

Solution of the Week #567 - Basketball Tournament

The most a team could play is 12, if they played all of the other six teams twice each.

The fewest possible is 6, if they played each of the other teams once. So on the face of it it looks as if the seven different possible numbers of games are 6,7,8,9,10,11,12.

However if a team played each other team once there can’t be team that played every team twice, as that would include the first team, leading to a contradiction.

So there are two scenarios regarding number of games played: either 6,7,8,9,10,11 or 7,8,9,10,11,12. If you were expecting seven different possibilities, reread the question and find that only the other six team’s numbers of games have to be different, and Nuneaton Predators number of games can be a repeat of one of those. That is the key to finding the number of games played by the Predators.

Let’s just consider the 7,8,9,10,11,12 scenario. One team, let’s call it A, must play every team twice, to become a team with 12 games. Another team, say B, plays A twice and every other team once, totalling 7 games. Team C plays A twice and B once, and every other team twice to be the team with 11 games. Team D plays teams A and C twice each, and team B and every other team only once, to total 8. Team E plays B and D once each and every other team twice, making 10. Teams F and G have played A, C and E twice, and B and D once, making 8 so far. If they play each other once, they will both be on 9 games, but if they play each other twice they will both be on 10 games, along with team E. In that second situation, no team has played exactly 9 times, so that can’t be the correct situation. Those two teams play 9 games each, and one them must be Nuneaton Predators.

If you do the same thing considering the 6,7,8,9,10,11 scenario, you will again find that the repeated number of games is 9, and so one of those teams must be Nuneaton Predators.

Nuneaton Predators played nine times.

Solution of the Week #566 - Central Area

No matter the proportions of the rectangle, it can be stretched by some factor in the horizontal and by the reciprocal of that same factor in the vertical, preserving the are of each of the regions. That being the case we can assume the rectangle is a square.

The upper right region will have sides of sqrt(2). The other two area 1 regions will have sides of ‘s’ and ‘s’-sqrt(2). We can therefore show that the side length ‘s’ must be equal to (sqrt(2)+sqrt(10))/2, or sqrt(2) times the golden ration.

The overall area of the square will then be sqrt(5)+3, and so the central area is sqrt(5).

Solution of the Week #564 - Comparea

If we assume the square is a unit square, then the triangular region has area 1/2.

The radius of the large circle is sqrt(2)/2

The radius of the smaller circle is 2-sqrt(2).

The area of the crescent region is therefore: (4*sqrt(2)-11/2)*pi, or about 0.493.

Therefore the triangle area is very slightly larger than the crescent area.

Solution of the Week #563 - Powers

a+b = s

a*b = p

What is a^4 + b^4 in terms of s and p?

 

Let’s start by squaring (a+b):

s^2 = a^2 + 2ab + b^2

Let’s isolate (a^2+b^2)

(a^2+b^2) = s^2 – 2p

Let’s square both sides:

a^4 + 2a^2b^2 + b^4 = s^4 – 4ps^2 + 4p^2

Again let’s isolate (a^4+b^4)

a^4 + b^4 = s^4 – 4ps^2 + 4p^2 – 2p^2

 

a^4 + b^4 = s^4 – 4ps^2 + 2p^2

 

And of course this works, and is real, even if the choice of s and p mean that a and b must be complex. For instance if the sum and product were 1 and 2 respectively, the sum of the fourth powers would be, using the formula, 1. But if you calculate a and b you would find messy expressions involving i and the square root of 7.

Solution of the Week #562 - Zig Zag Sequence

We saw previously how a related sequence 1,2,2,3,3,3,4,4,4,4 etc could be expressed as a square root rounded to the nearest integer.

a(n)= round(sqrt(2*n))

Next I’m going to square each term in that sequence:

1,4,4,9,9,9,16,16,16,16,etc.

b(n)= (a(n))^2

Our target sequence 1,3,2,6,5,4,10,9,8,7,etc. Can be expressed in terms of n and b(n):

c(n)= b(n)+1-n

For instance, the fourth term will be 9+1-4=6.

So the full expression will be:

c(n) = (round(sqrt(2*n)))^2 + 1 – n

 

If you were curious about what, say, the millionth term would be:

C(1000000) = (1414)^2 + 1 – 1000000 = 999397

Solution of the Week #560 - Concentric Circles

You could of course, assuming the actual radii are not important, let the smallest circle vanish to a point, at which point it becomes clear that the radii of the other two circles which must be 65 and 33, become the hypotenuse and one leg of a right triangle, with x as the other leg. x is therefore 56.

If you want to be more rigorous, let the radii be a<b<c.

33^2 = b^2 – a^2

65^2 = c^2 – a^2

x^2 = c^2 – b^2

65^2 – 33^2 = c^2 – a^2 – b^2 +2^2 = c^2 – b^2

x^2 = 65^2 – 33^2

x = 56

Solution of the Week #559 - Minimal Combinations

(2k)! must have at least three more factors of 5 within it that twice k! (the same goes for factors of 2 as well, but that will be achieved much earlier).

If k=63, 63 has 14 factors of 5 (floor(63/5) + floor(63/25)).

2*63 = 126, 126 has 31 factors of 5 (floor(126/55) + floor(126/25) + floor (126/125)).

For any smaller k (2k)!/(2*k!) is less than 3.

Solution of the Week #558 - Isosceles Trapezoid

The area of the lower region is 1/2*54*d

The area of the upper region is 1/2*30*e

Since those areas are equal we know that e/d = 54/30 = 9/5

Since the triangles with hypotenuse a and 13 are similar we know that a/13 is also 9/5.

a is therefore 23.4.

If the lower area is x, then the total area below the dashed line is 9x/5.

Since the total area must be 3x, the area above the dashed line must be 6x/5. It follows from this that b/30 must be 6/5.

b is therefore 36.

Since 54 = 36 + 2c, c=9.

To find e we use Pythagoras on the ace triangle. e = 21.6.

The total area of the trapezoid is therefore 21.6*(54+36)/2 = 972

Solution of the Week #556 - Triangle and Semicircle

If we call the radius ‘r’ and the height from the centre of the circle to the diagonal line ‘a’, then by Pythagoras we have:

49^2 = r^2 + a^2

Using intersecting chords theorem, 49*31 = (r+a)(r-a)

49*31 = r^2 – a^2

 If we take the first equation and subtract the second we get:

 49(49-31) = 2a^2

a^2 = 49*9

a = 21

 The height of the triangle is twice this distance ‘a’, and is therefore 42.