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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!
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.
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.
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.
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.
Because of the symmetry of the set-up we are able to determine a few facts, irrespective of the particular shape of the triangle. We are able to assign areas A, B, C, D to different regions as shown. From the triangle SUY, A+B+C+D=1.
We can identify similar triangles PQR and PST, whose linear scaling is 1:2 and so whose area scaling is 1:4. So:
4B = A+2B+C
Similarly PUV has an area nine times that of PQR, so:
9B = 2A+2C+3B+2D
The first leads to 2B = A+C and the second leads to 3B = A+C+D, therefore B=D. Then, since B+D = A+C and all four add to 1, it follows that B and D are each equal to 1/4.
Next, look at similar triangles STX and QRW. The former is twice the linear scale, so four times the area of the latter, so:
C+D = 4C
3C = D
But we know that D=1/4, so C=1/12.
Since A+B+C+D=1, A = 1-1/4-1/12-1/4 = 5/12.
The area A is equal to five twelfths.
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.
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.
CHEF CAME TO IT BUT NOT WISE AT BRAG AT OUR ASH AS FAST ONE
becomes:
THEF LAME TH AT BUR NST WICE AS BRIG HT BUR NSH AL FASL ONG
then:
THE FLAME THAT BURNS TWICE AS BRIGHT BURNS HALF AS LONG
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
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.
The more obvious solution is that 2 x 4 x 6 x 8 = 384.
The second is a little bit sneaky, as I didn’t specifically mention you could use negative numbers (as long as the result is positive): -7 x -1 x 5 x 11 = 385.
Things work out best for those who make the best of the way things work out. John Wooden.
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.
You can avoid reality, but you can't avoid the consequences of avoiding reality.
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.
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
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).
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%.
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.