Solution of the Week #326 - Inch Centimetre Prime

To be a whole number of centimetres, the number of inches must be a multiple of 50. For the number of centimetres to be even, the number of inches can be further narrowed down to multiples of 100. For both the number of inches and the number of centimetres to avoid being one more than a multiple of 3, the number of inches must be a multiple of 300. This straight away narrows down the task to the point where we can check individual cases.

If the number of inches were 2100, that would convert to 5334cm: 2099, 5333 and 7433 are all prime.

So the distance is 2100 inches or 5334 cm.

This is the only possible distance under 200m as the next set of numbers that fit the criteria are: 9000 inches (22860cm).

Solution of the Week #325 - Uprooted

Let √x=c

Let √y=d

The two given equations become:

c^2d+cd^2=14

c^3+d^3=22

If you cube (c+d), you will get an expression purely based on the two above equations:

(c+d)^3 = c^3+d^3+3c^2d+3cd^2

(c+d)^3 = 22 + 14*3

(c+d)^3 = 64

(c+d) = 4

Since the first equation can also be written as cd(c+d) = 14,

cd = 14/4 = 3.5

If you square (c+d) you get:

(c+d)^2 = c^2 + d^2 + 2cd

But c^2 = x and d^2 = y, and we know (c+d) and cd, so therefore we can rearrange to:

(x+y) = (c+d)^2 - 2cd = 16 - 7

(x+y) = 9

Solution of the Week #324 - Number Hunt

5623109 = 23 x 41 x 67 x 89

There are in total 12 ways of combining four two-digit primes with no digits in common, but only one of the twelve products contains no repeating digits.

This is how you can reduce the number of possible arrangements to just twelve (which can then be easily checked to see if the product contains no repeated digits), just with a little logic:

Once we are past single digit primes, all primes must end in 1, 3, 7, or 9. Since we are after four primes with no repeated digits, they must end in these four digits, and therefore they cannot start with those four digits. They also cannot start with a 0 as they wouldn’t be two-digit numbers, so the start numbers must be four of the following five: 2,4,5,6,8. Since the primes starting with 2,5,8 all end in 3 or 9, only two of those three start numbers will be used, and the 3 and 9 will also come from those. Therefore 4 and 6 must be in any selection, and must be followed by 1 and 7.

In summary, two of the numbers must start with 2,5,8 and end with 3,9 (6 possibilities), and two must start with 4,6 and end with 1,7 (2 possibilities). Overall there are therefore 6x2= 12 possibilities.

Solution of the Week #323 - Triangle Rectangle 2

You can remove the line CD as it is not really helping, and instead introduce another line, parallel to CD, but passing through F.

On the face of it, there are too many degrees of freedom to be able to determine anything useful. HOWEVER, since we want DF to be a minimum, we can also add in the fact that angle EFD must be a right angle, and that lengths x and y must be equal.

Looking at the three similar triangles around the outside we can tell that:

64/x = y/z = w/v

As well as the facts that:

x = y; z + w = 64; 2x - v = 75

We can either do a bit of algebra and find a cubic in terms of x, which factorises to one real and two complex solutions.

Or we can try a few values for x and see what works. It turns out that x=y=48, z=36, w=28, v=21, and therefore DF = 35.

Solution of the Week #322 - Number Hunt

Since p-1 will be even for any odd prime, 2 must be a prime factor of n, and as it is also one less than a prime, so is 3.

Since 2x3 is one less than a prime, 7 is also a prime factor of n. Of the products of subsets of 2,3,7, only 2x3x7 is one less than a prime, so 43 is also a prime factor of n.

Of the products of subsets of 2,3,7,43, none are one less than a prime, so we can stop there.

n = 2x3x7x43 = 1806.

Solution of the Week #321 - Missing Digits

7, 1 and 1.

2 4/7 x 3 1/9 x 1 1/4 = 10

Solution of the Week #320 - Triangle and Rectangle

The minimum length of DF is 39.

Solution of the Week #319

The answer is 16

w^5 + x^5 + y^5 + z^5 = -8

w^6 + x^6 + y^6 + z^6 = 2

w^7 + x^7 + y^7 + z^7 = 16

The values of w,x,y,z in some order are 1, -1, 1+i, 1-i.

Solution of the Week #318

ANSWER = 128

The highest prime factor of 128 is a=2,

The highest prime factor of 126 is b=7,

The highest prime factor of 119 is c=17.

Solution of the Week #317

Because of similar triangles, the two ships will each reach their destination at the same time.

Solution of the Week #315 - Shaded Area

If you draw a vertical line between the apexes of the two triangles you will exactly divide the shaded area in two. Each half will have a side of 1 and another of x, and also a fixed angle (opposite the side 1) of 150 degrees. In order to maximise the area of this triangle, we must maximise the distance between the vertex with angle 150, and the side of length 1. This will be achieved when the triangle is isosceles, such that the distance between the two apexes is also x.

Since cosine of 150 degrees is -sqrt(3)/2, it’s possible to calculate x to be sqrt(2-sqrt(3)), which is approximately 0.5176.

Solution of the Week #314 - Digit Sums

The longest possible streak is 78.

The first such streak starts with 9,999,999,961 and continues until 10,000,000,038.

Solution of the Week #313 - Alien Number System

13041, which in our number system is 3x3x3x3x7x23, in the special number system can either be 21x621 or 81x161, all four of which are prime in the special system (but none prime in real life).

Solution of the Week #312 - Wolf Tooth

Below is a diagram showing the necessary arrangement of the colours. The cube colours are in the squares, and the octahedron colours are in the hexagons.

The colours on the octahedron vertex in the middle of the green cube face are: Red, Orange, Silver, Blue.

And here is an image of the final cube, courtesy of Philip Morris Jones

Solution of the Week #311 - Diamond Ring

If we draw a vertical line through the middle of the figure it should be clear that the diameter of the large circle is the sum of the diameters of the other two circles. It should also be clear that this vertical line will pass through the midpoint of BD. Since we are told ABCD has side length of 2, BD has length sqrt(8) and each half of BD has length sqrt(2).

Using the intersecting chords theorem, we know that the product of the two smaller diameters must be the same as the product of the two halves of BD, namely 2.

If we call the RADIUS of the smallest circle x and the radius of the other inner circle y, then we know that 2x*2y =2, and so xy=1/2.

The outer circle has radius z = x+y

To find the shaded region we simply need to find the area of the outer circle and subtract the areas of the other two circles.

Shaded region S = (pi)z^2 – (pi)x^2 – (pi)y^2.

S/pi = z^2 – x^2 – y^2

= (x+y)^2 – x^2 – y^2

= x^2 + 2xy + y^2 – x^2 – y^2

= 2xy

But we know xy=1/2, and so S/pi = 1.

Therefore the shaded region S = pi, and is not dependent upon the size of the circles.

Solution of the Week #310 - Always True?

Yes, it’s always true.

 If we call our first number a^2+b^2 and our second number c^2+d^2, since we are told that each is the sum of different squares we can, without loss of generality, specify that a>b and c>d. It is possible for a=c or for b=d, but not both at the same time, since we are told the two sums of squares are different.

If we multiply out (a^2+b^2)(c^2+d^2), we get (ac)^2+(ad)^2+(bc)^2+(bd)^2, which is the sum of four squares, not what we are after. Our approach will be to call the target product (e^2+f^2) and to express e and f in terms of a,b,c,d and demonstrate that (e^2+f^2) is equivalent to (ac)^2+(ad)^2+(bc)^2+(bd)^2.

 This can be achieved by letting e=ad+bc and f=ac-bd

 e^2+f^2 = (ad)^2+2abcd+(bc)^2+(ac)^2-2abcd+(bd)^2

Clearly the abcd terms cancel out, leaving only the terms we were hoping for.

 However, this isn’t quite enough to satisfy the question, since e^2 and f^2 need to be different, and our expressions do not guarantee that. For instance if, as in the original example, a=2, b=1, c=3, d=1, both e and f are equal to 5.

 Instead let us call g=ac+bd and h=ad-bc (whereas before f was guaranteed to be positive, h now could be negative, but it doesn’t really matter, as h^2, which is what we are really interested in, will be positive regardless).

 Now g^2+h^2 = (ac)^2+2abcd+(bd)^2+(ad)^2-2abcd+(bc)^2

Again the abcd terms cancel and leave just what we want. But now, since a>b and c>d, g^2 will always be bigger than h^2, meaning we always have two different squares summing to the target value. This method might yield h^2 = 0, which is technically a square number, but if we would prefer non-zero square numbers we can revert to the e and f equations, knowing that if h = 0, e cannot be equal to f. I have a nice proof of this, but I won’t go into it now.

Solution of the Week #309 - Three Fences

The greatest area bounded by four straight edges occurs when those edges form a cyclic quadrilateral (all four vertices lie on a circle). Given that we can choose the length of wall that we use, it makes send for that to be the diameter of the circle, with length 2r. Then the three fences each form isosceles triangles with r as the common lengths and 25, 33 and 39 as the ‘bases’. It turns out that for this to happen, r=32.5, and the three triangles have ‘heights’ of 30, 28 and 26 respectively. The area is therefore 1344 square metres.

Solution of the Week #307 - Descend the Ladder

For a number to directly reach ‘n’, a number ‘m’ must exist that is the highest prime factor of n+m+1. If n+m+1 is divisible by m, which is a necessary (but not sufficient) condition for this, then n+1 must also be divisible by m. So we can find all possible candidates for numbers that can directly lead to 5 by looking at the prime factors of 5+1. These are 2 and 3.

Considering m=2, our n+m+1 is 8, and 2 is indeed the highest prime factor. So 8 leads to 5.

Considering m=3, our n+m+1 is 9, and 3 is indeed the highest prime factor. So 9 leads to 5.

Now let’s consider n=8, the only prime factor of n+1 is 3, so 12 is a possible candidate we need to check. The highest prime factor of 12 is 3, so that works.

Next, n=9 gives us the candidates 12 again, and 15. 15 works as its highest prime factor is 5.

Consider n=12, the only prime factor of n+1 is 13, giving us 26, which works.

For n=15, m=2, the candidate of 18 comes up, however the highest prime factor of 18 is 3, so this doesn’t work and proves a dead end.

Finally, n=26, m=3, we need to check to see if the highest prime factor of 30 is 3. It isn’t, so this is another dead end.

So the full list of composite numbers that reach 5 is: 8, 9, 12, 15 and 26.