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.

three fences soln.jpg

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.

Solution of the Week #306 - Climb the Ladder

38 +19+1

58 +29+1

88 +11+1

100 +5+1

106 +53+1

160 +5+1

(It’s about this point you start to think you’ll never hit an odd number, let alone a prime number)

166 +83+1

250 +5+1

256 +2+1

(but by landing on a power of 2 we now change parity)

259 +37+1

297 +11+1

309 +103+1

413 +59+1

473 +43+1

517 +47+1

565 +113+1

679 +97+1

777 +37+1

815 +163+1

979 +89+1

1069, which is prime!

Solution of the Week #305 - Power Tower

If you try to calculate the number you won’t get very far. 5^7^5 is already way too large for a calculator, at over 11000 digits. The number of digits in the entire number is unimaginably large. But of course we only need the final five.

If we look at the pattern of the final five digits of 5^n, (also known as the remainder of 5^n when divided by 100000), we see a nicely repeating pattern. Once the numbers are big enough, the final five digits cycle around just eight options.

So to find the last five digits of 5^(very large number), we only need to know the remainder of the (very large number) when divided by 8.

But the (very large number) we are concerned with is equal to 7^(quite large number), and the remainder of 7^n when divided by 8 alternates between just two numbers: 1 when (quite large number) is even and 7 when (quite large number) is odd.

But the (quite large number) is equal to 5^(large number). 5 to the power of any whole number will be odd, since 5 is odd. Therefore 7^(quite large number) gives a remainder of 7 when divided by 8. Therefore the final five digits of 5^(very large number) are 78125.

Solution of the Week #304 - Compass Points

The fact that opposite angles of the quadrilateral add to 180 degrees tells us that the quadrilateral is cyclic, ie that a circle drawn through any three vertices will also pass through the fourth.

The fact that the four vertices can lie on a circle lets us use a fact about lines crossing in circles: the product of two opposing distances from a particular point within a circle to the edge of the circle will always be the same, regardless of the angle of the line through the point.

In the following diagram for instance, ab = cd, as a general rule.

compass points sol.JPG

We can use this fact if we take the crossing points as the centre of the circle, and the four distances as (12+r), (31+r), (27+r) and (10+r).

(12+r)(27+r) = (10+r)(31+r)

324 + 39r + r^2 = 310 + 41r + r^2

14 = 2r

r=7

So the radius of the circle is 7.

Solution of the Week #303 - Three Tangent Circles

I deliberately drew the original diagram misleadingly so to not give the game away, but the fact that AB and CD are equal means that EF and GH are also equal, and that a line drawn between the centres of the top and right circles will be at 45 degrees downwards. It also follows that the line FG will be tangent to the top and right circles, and so will be equal in length to the line joining their centres. Since this is simply twice the unit radius, the length of FG will be 2.

THREE TANGENT CIRCLES sol.JPG

Solution of the Week #301 - Circular Table

Since every six consecutive friends adds to the same amount, it follow that 1-6 totals the same as 2-7, and since these two groups have five people in common it follows that 1 is the same age as 7. By the same reasoning, every person is the same age as the person 6 seats away. So 1, 7, 13 and 19 are all the same age. But the group (19,20,21,1,2,3) has the same total as (20,21,1,2,3,4), so 19 and 4 are the same age. Following this through the ages repeat in a pattern every three people, and each consecutive group of three people must add up to 100.

We are given the ages of 1 and 8 and asked for the age of 15. But 8 is the same age as 2, and 15 is the same age as 3, and together they will form a consecutive group of three adding to 100. If the first two are aged 25 and 33, the third must be aged 42.

The person seated at position 15 is aged 42.

Solution of the Week #300 - Winner Stays On

At first sight the question seems impossible. There is definitely not enough information to decide who plays who and when but, believe it or not, there in enough information to answer the particular question of who lost in the ninth game.

Firstly lets see how many games there were altogether. To do this we an simply add together the number of games played by each individual, and divide that total by two, since there are two players in each game. This tells us that there were 11 games in total. Because of the way they have of rotating players, even if a player loses, they will be back at the table three games later. The key now is to look at Billie, who has played the least games. If we assume they lost every game they played they would sill play roughly a third of the games. Say Billie played in game 1, they would play again in game 4, game 7 and game 10, but that is too many games. Similarly if they played and lost in game 2, they would reappear for game 5, game 8 and game 11. Still too many. The only possibility is that Billie played and lost in game 3, game 6 and game 9.

Billie lost in the ninth game.

Solution of the Week #298 - Unit Fractions

The numbers in the denominators are exactly those numbers that have either 2 or 5 or both as their prime factors, eg 8 = 2x2x2, 10=2x5. This is exactly the property which means they have a terminating decimal expansion.

You can organise the factions into a different order such that you first add those with no 5 factor at all:

A = (1 + 1/2 + 1/4 + 1/8 + 1/16 + …)

If you first double it:

2A = (2 + 1 + 1/2 + 1/4 + 1/8 + 1/16 + …)

And then you subtract the original:

A = 2

Next you look at those fractions which have exactly 1 factor of 5:

(1/5 + 1/10 + 1/20 + 1/40 + 1/80 + …)

You can factorise the 1/5 out of each term:

1/5 x (1 + 1/2 + 1/4 + 1/8 + 1/16 + …)

But that is just: 1/5 x A

The next group will be fractions with exactly 2 factors of 5, and similarly will total to 1/25 x A

The entire sequence will total:

AB, where B = (1 + 1/5 + 1/25 + 1/125 + …)

Multiply B by 5:

5B = (5 + 1 + 1/5 + 1/25 + 1/125 + …)

And subtract B:

4B = 5, so B = 5/4

The overall sum AB is therefore 2 x 5/4 = 5/2 = 2.5


Solution of the Week #297 - 2021 Puzzle

AB^4+CD^4 = 2021

AB^3+CD^3 = 485

AB^2+CD^2 = 101

AB+CD = 5

 

If you were to multiply one of the equations throughout by (B + D) you will get an equation involving some of the other terms on the left-hand side, along with BD and (B+D), for instance:

 

AB^3+CD^3 = 485, multiplied by (B+D):

 

1) AB^4+CD^4 + BD(AB^2+CD^2) = 485(B+D)

 

AB^2+CD^2 = 101, multiplied by (B+D):

 

2) AB^3+CD^3 + BD(AB+CD) = 101(B+D)

 

AB+CD = 5, multiplied by (B+D):

 

3) AB^2+CD^2 + BD(A+C)= 5(B+D)

 

In each of the equations let (B+D)=E and BD=F, and replace each of the ABCD terms with their numerical value, if known:

 

1) 2021 + 101F = 485E

2) 485 + 5F = 101E

3) 101 + (A+C)F = 5E

 

Multiplying 1) by 101 and 2) by 485 and subtracting one from the other gives an equation involving large numbers and F, but neatly divides to give F = 4. Plugging that back in to, say 2), gives a value of E = 5. Plugging both of these values into 3) gives a value of -19 for (A+C).

To find B and D we need to find a pair of numbers whose sum is 5 and whose product is 4. These must be 1 and 4. Let us assume for a moment that B is 4 and D is 1.

So we know that A+C is -19 and 4A+C is 5. Therefore A = 8 and C = -27.

(If we were to have assumed that B is 1 and D is 4, we would get A = -27 and C = 8, so the same four numbers but in a different order.)

Since we are asked for ABCD we get the same answer either way: -864.

Solution of the Week #294 - Special K

The multiples of 11 will coincide whenever (K^2) – (4 x 5) is a multiple of 11. This occurs when K is either 3 more or 3 less than a multiple of 11, so 3, 8, 14, 19, 25, 30 etc.

The first three times this is a prime number are 3, 19 and 41.

Therefore the answer is 41.

For an explanation of why this is the case, read on:

Under the assumption that KM+5L and KL+4M are both congruent to 0, modulo 11, we can multiply either side by an integer and that congruence will be preserved. Therefore we can multiply the first expression by K (which is defined as an integer) and the second by 5. Therefore K^2.M+5LK and 5LK+20M are also both 0 mod 11.

Subtracting one expression from the other unifies the two expressions but maintains the divisibility by 11. Doing so makes the 5LK terms cancel out so we end up with (K^2 – 20)M is 0 mod 11. Therefore for our purposes, where we are interested in the special values of K, this occurs when K^2 – 20 is a multiple of 11, which is where we came in.

As a side note, if you’re wondering what happens if you follow the other strand and assume M is a multiple of 11, it’s not difficult to show that that means that L must also be a multiple of 11 for either/both of the original expressions to be a multiple of 11, which it will be for all K.