Solution of the Week #291 - Divisible by 32

n=2

Since 26 and 6 are both even, if n is greater than or equal to 5, both parts will independently be divisible by 32, so be only need to consider n = 1,2,3,4.

When n=1, it is obvious that 26+6=32.

When n=3, we note that (x^3 + y^3) can always be factorised into

(x+y)(x^2–xy+y^2), so letting x=26 and y=6 shows that this is divisible by 32.

When n=4, each part will be an odd number multiplied by 2^4, so will naturally have a remainder of 16 when divided by 32. Putting both of them together gives a multiple of 32.

That just leaves n=2. Unlike all of the other numbers, there is no inherent reason why this would be divisible by 32, and simply checking shows that it isn’t.

Solution of the Week #290 - Special Sequence

One possible answer is:

9 10 11 12 4 5 6 7 8 4 9 5 10 6 11 7 12 8

The structure hints at a general method (which might not have been apparent from the original example, but nevertheless follows the same idea). Using numbers m to n, n needs to be (at least) 3m.

If you’re interested I developed this into a sequence which has now been published on the OEIS (Online Encyclopedia of Integer Sequences): https://oeis.org/A338804

Solution of the Week #289 - Power Play

The remainder is zero. In fact 10^n + 11^n is divisible by 21 whenever n is an odd number.

You can prove this by showing that when n is odd, (a + b) is a factor of (a^n + b^n).

You can factorise (a^n + b^n) as:

(a + b)(a^(n-1) – a^(n-2)b + a^(n-3)b^2 - … - ab^(n-2) + b^(n-1))

When you multiply the massive second bracket by (a + b) almost everything cancels out, leaving only (a^n + b^n). Here is a specific example using n=5:

(a+b)(a^4 – a^3.b + a^2.b^2 – a.b^3 + b^4)

Multiplying each term in the second bracket first by a:

(a^5 – a^4.b + a^3.b^2 – a^2.b^3 + a.b^4)

And then by b:

(a^4.b – a^3.b^2 + a^2.b^3 – a.b^4 + b^5)

Adding those two brackets leaves only the first and last terms:

(a^5 + b^5)

So as long as a, b and n and integers, the whole second bracket will be too. Therefore (10 + 11) divides into (10^101 + 11^101).

Solution of the Week #288 - Round Table

There are 64 (2^6) possible gender permutations of the six children. For all the males to be seated together, all of the girls need to be older than all of the boys, and there is one such arrangement for each of the possible numbers of girls from 0 to 6, giving 7 possible valid arrangements.

Therefore the probability is 7/64, which is just under 11%.

Solution of the Week #287 - Divided Rectangle

Call the missing area x^2.

Let the base of the rectangle equal 2x/c (where x is positive, c is just an arbitrary scaling factor).

Since the area of a triangle is half base times height, the height of the missing triangle is xc.

Since the 4-triangle is similar to the missing triangle, the height of that is 2c (as 2^2=4), so the overall height of the rectangle is (xc+2c).

The area of the overall rectangle can now be written is two distinct ways, which must be equal:

Summing the four regions: x^2+15+4+26

Rectangle base times height: 2x^2+4x (notice the c vanishes)

Equating those two expressions and bringing everything to one side gives:

x^2+4x-45=0

which we can factorise into:

(x+9)(x-5)=0

The positive root is x=5, so the missing area x^2 is 25.

Solution of the Week #286 - The Cube and the Cup

To keep things relatively simple I will assume the circular rim of the cup has a radius of 1.

It’s not important to solving the question, but for the two orientations to be possible, the edge length of the cube must be somewhere between sqrt(6)/2 (1.225..) and 2. Any smaller than that and the point-down cube would fall in, any larger and the edge-down cube would not fit within the rim.

So on the assumption that we are within that range and both configurations are possible, it turns out that for the point-down orientation, the cube side length doesn’t matter: in theory you could have a mile long cube and the portion below the rim would be exactly the same.

To work out what that depth is for our unit radius cup, consider a cross-section through the cube and cup through point A O and D:

Since the entire face that triangle BCO is part of is perpendicular to edge AO, so is line OD. Projecting down to the side view we can use the fact that an angle in a semicircle is a right angle to ascertain that the depth is sqrt(2)/2 or 0.707…

So now we need to find the cube size for which the edge-down cube will also be at that same depth. By looking side-on at the diamond-shaped end of the cube with the required depth, and then looking at the plan view, we can see that the edge length of the cube must be sqrt(2) = 1.414…. Inevitably, this lies in our range of acceptable cube sizes.

So finally, how can you tell simply and visually if the edge length of the cube (as a ratio of the radius of the cup) exceeds sqrt(2)? Do you remember the forgotten third orientation – face-down? Simply if the face-down cube lies within the rim of the cup, then the ratio of edge length to radius is less than sqrt(2) and the edge-down orientation will be deeper than point-down, whereas if the face-down cube can sit on top of the cup with all four corners outside of the rim, then the point-down orientation will be deeper.

Solution of the Week #285 - Left Right Pairs

There are twelve different numbers in the list, so logically it can’t be done in fewer than 7 parts (3 of one hand and four of the other, to combine in 3 x 4 = 12 different ways).

However the fact that 23 is the only odd number in the list throws a spanner in the works as it effectively means one part will help make up the 23 but not be used elsewhere. Assuming that you can find a way of making the other 11 with just 3 + 4 distinct parts, 8 parts in total will be required.

If we assume for the moment that there are only 3 L parts, and that one of them has length 0 (impossible ultimately, but useful right now), then what we are searching for is a subset of four of the 11 lengths (so disregarding 23), and two number a and b such that the remaining 7 numbers can be made by adding either a or b to one of the 4 numbers.

By experimenting, the subset 20, 44, 48 and 50, with a and b values 8 and 26 fits the bill. To avoid the zero length part, we can simply subtract 10 from each of (20,44,48,50) and call those the L parts, and add 10 to (0,8,26) to find the R parts, adding in an extra R part to make 23 possible. Thus:

20 = 10L + 10R

23 = 10L + 13R

28 = 10L + 18R

44 = 34L + 10R

46 = 10L + 36R

48 = 38L + 10R

50 = 40L + 10R

52 = 34L + 18R

56 = 38L + 18R

58 = 40L + 18R

70 = 34L + 36R

74 = 38L + 36R

which uses the following eight parts: 10L, 34L, 38L, 40L, 10R, 13R, 18R, 36R.

Solution of the Week #284 - Quadrilateral in a Circle

Area = 936.

To work through the problem in trigonometry is a bit of a nightmare so I’d envisage people solving the problem by drawing lines from all vertices and midpoints to the centre of the circle, calculating the cosine of each of the angles at the centre in terms of ‘r’, and then varying the value of r until the angles add up to 180 degrees. This occurs when r is precisely 25. From there we can easily work out the side lengths of all 8 of the internal right-angled triangles and tot up their areas to get 936.

If you do that you might be surprised to discover that the eight triangles are two sets of four congruent triangles: (7,24,25) and (15,20,25).

Solution of the Week #283 - Number Hunt

A, B, C and D are 3, 4, 5 and 2 respectively, since 34^5 = 45435424

Solution of the Week #282 - Heptagon Revisited

Using the fact that an angle in a semi-circle is always a right angle, we can see why the given circle passes through the midpoints shown. Similarly we can see why the smaller circle in the second figure also passes through midpoints of the heptagon. Now we only need to find the diameter of that circle. I only give you an overview of the solution for you to fill in the gaps if you desire.

We can find angle ACB easily enough, and since triangle ABC is isosceles, we can find the length of AC.

Triangle ABC is similar to triangle ACD, therefore AB:AC = AC:CD, so we know length CD.

By examining the angles triangle CDE is isosceles, therefore CE = CD.

Since the heptagon is regular, we know the angles within isosceles triangle CEF, so we can know the length of CF.

Moving to the next figure, GH = CE, and IJ = CF.

GHIJ and IJKL are similar trapezoids (sides and diagonals parallel between the two figures) therefore GH:IJ = IJ:KL.

Therefore if we know AB = 6296, we also know that KL = 1247 (to the nearest integer, 1247.000015… to a greater degree of accuracy).

If you’re interested, the exact ratio between AB and KL is:

(1+2sin(270/7))^2

Solution of the Week #281 - Unit Fractions

Since both 8 and 188 divide evenly into 43992, the original sum of three fractions can be written as (5499+234+1)/43992. This further cancels to 61/468. So we have:

61/468 = 1/a + 1/b         multiply each term by 468ab to eliminate denominators:

61ab = 468b + 468a         multiply each term by 61, and move everything over to the left:

61x61ab – 61x468a – 61x468b = 0      add 468x468 to both sides:

61x61ab – 61x468a – 61x468b + 468x468 = 468x468

So far we just seem to be needlessly complicating it, but in fact we’ve been leading up to the next step, which is to factorise the left hand side:

(61a-468)x(61b-468) = 468x468

Now here’s the really clever bit. Since a and b have to be whole numbers, the remainder when divisible by 61 of (61a-468) or (61b-468) will remain the same regardless of the value of a and b. Their ‘modulo 61’ value will each be 20. The point of doing this is that however we factorise the right hand side, the two parts that correspond to (61a-468) and (61b-468) must also have a value of 20, modulo 61.

468x468 = 219024 = 2x2x2x2x3x3x3x3x13x13, so there are plenty of options for pairs of numbers that multiply to make 219024:

1x219024, 2x109512, 3x73008, 4x54756, 6x36504, 8x27378, 9x24336, 12x18252, 13x16848, 16x13689, 18x12168, 24x9126, 26x8424, 27x8112, 36x6084, 39x5616, 48x4563, 52x4212, 54x4056, 72x3042, 78x2808, 81x2704, 104x2106, 108x2028, 117x1872, 144x1521, 156x1404, 162x1352, 169x1296, 208x1053, 216x1014, 234x936, 312x702, 324x676, 338x648, 351x624, 432x507, 468x468,

However we are only interested in numbers that are equal to 20 mod 61. 81 is such a number, and its complement 2704 is the only other.

So we can let 61a-468=81 and 61b-468=2704 (or vice versa). These solve simply to a=9 and b=52 (or vice versa), and so these are the only solutions to the puzzle.

To conclude:

1/8 + 1/188 + 1/43992 = 1/9 + 1/52

Solution of the Week #280 - Sum of Cubes

There are infinitely many solutions but the smallest is that a, b, c and d are (in some order) -11,-2,7 and 10.

Their sum is 4, and the sum of their cubes (-1331, -8, 343, 1000) is also 4.

I arrived at this answer partly by narrowing the scope of where I was looking, and thereafter trial and inspection. I remembered that ALL cube numbers are either 0,1 or 8 modulo 9 (in other words they all cube numbers are either a multiple of 9, or one away either side). You can verify this astonishing fact by just trying the first nine cube numbers and observing that this is in fact true for those, and because we are working in module 9, it is therefore true for all cube numbers.

For our four cube numbers to add to 4, they must all be of the +1 variety. This happens whenever the number to be cubed is 1 more than a multiple of 3, ie: 1, 4, 7, 10, etc and -2, -5, -8, -11, etc. So we only need to look at the cubes of these numbers. Obviously the trivial 1 appears in this list, as do each of the numbers in my answer: -11, -2, 7 and 10.

Solution of the Week #269 - Heptagon Figure

The angles marked in this figure are in units of 180/7°. So the internal angles in a heptagon will add up to 35, the internal angles in a pentagon will add up to 21, in a quadrilateral will sum to 14 , and in a triangle or on a line will sum to 7.

By definition all of the angles in the regular heptagon will be 5. From pentagon NBCDE, angle ENB is 1 (same for angle GPD).

Since H and K are midpoints of the heptagon, FGHK is a trapezoid, therefore HK is parallel to GF, and so angle HKE is 5 (same for JLD). The supplementary angles MKN and MLP are therefore 2.

From the inherent symmetry of the figure, and the angles in the quadrilateral NPDE, angles ENP and DPN are each 2. Therefore from triangle NMK, angle NMK is 3 (same for PML). From the pentagon MKEDL, angle KML is 1.

Finally, since NMK is isosceles, NM = MK (same for MP). So if MK is 7, NP = 14.

Solution of the Week #268 - Dice Game

It should come as little surprise that winning the first round gives you an advantage as the game continues. You are already more likely to win the second round, and if you do your advantage is further increased.

It is however by no means a foregone conclusion. Examining all of the possibilities the probability of the winner of the first round going on to win works out at about 63.5%, with the chance of the initial underdog seeing a reversal of fortune as the remaining 36.5%.

The exact values, should you be interested, are as follows:

Initial winner winning: 137009/215622

Initial loser winning: 78613/215622

The most likely specific sequence of results by a long way is for the initial winner to simply win every round. This occurs with a probability of 3315/10648 or just over 31% and so alone accounts for almost half of the initial winner’s possible routes to victory.

Solution of the Week #267 - Tetrahedron Game

As it rolls around the grid, the tetrahedron will always present the same face to particular triangle of the grid. In fact the grid can be subdivided into four classes of triangles, each of which can only be landed on by one particular face of the tetrahedron. That being the case, a selection of painted triangles is solvable if and only if it contains one each of the four classes of triangles. Grid 3 is the only such grid.

Interestingly, so long as the grid is solvable, it doesn’t matter where you initially place the tetrahedron, and you can also decide to end up on any particular triangle of the grid (even if it isn’t initially painted).

Solutions

As you know if you follow this site, for a few years I didn’t publish the solutions to my Puzzles of the Week, but starting this year I have published a solution each Monday for the puzzle of the previous Friday. This has proved popular and I’ve found in writing them that the solution is often more interesting than the puzzle itself. I’m very pleased that I made the decision to start including solutions, but I only wish I’d started earlier.

It is my intention to eventually fill in the gaps and provide solutions for all the previous puzzle, Squarespace allows me to back-date them so that they will slot in between the puzzles.

With currently 232 solutions missing, this is not a quick task. If there are any old puzzles you would like me to provide a solution for, just let me know and I’ll be very happy to bring them to the top of the to do list.

Solution of the Week #266 - Base Four Code

Anyone who stops learning is old, whether twenty or eighty. Anyone who keeps learning stays young. The greatest thing you can do is keep your mind young.

This has been variously attributed to Henry Ford and Mark Twain.

Solution of the Week #264 - What is the Angle?

Here are a couple of different approaches:

Firstly I’m going to make a copy of triangle ABC, rotated 90 degrees anti-clockwise around point A to form triangle AB’C’.

Since ACEC’ is a 35 by 35 square, we can easily find out what DE and B’E are: 21 and 20 respectively. This forms a Pythagorean triple with the hypotenuse, B’D, equal to 29.

However since BD is also (14+15=) 29, AB equals AB’ and AD is shared by both, triangles ABD and AB’D are similar and angle x is equal to the angle we seek.

But since angle BAB’ = 90 degrees (as B’ was formed by rotating B by 90 degrees), the angle we seek must be half of this, namely 45 degrees.

An alternative approach is to scale down the two parts of the puzzle triangle as below. This can be done because it doesn’t change the angle we are seeking to find.

Similar to before we can rotate the left hand triangle 90 degrees anti-clockwise. Now to complete a rectangle we just need a copy of the 5x2 triangle, which means the triangular space in the middle is a right angled isosceles triangle, so x is 45 degrees. Since x + ? = 90 degrees, ? is also 45 degrees.

Solution of the Week #263 - 24-gon

That was where my original solution ended, however following feedback after the puzzle went live on Friday, it isn’t necessarily clear that three equally spaced segments have a constant combined sum, so I have added proof of that. Since the edge length of the 24-gon is the same we only need to consider the perpendicular distance to the internal point. Since this three segments are equally spaced, if we extended the bases we would form an equilateral triangle, making it equivalent to the situation illustrated below.