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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.
7, 1 and 1.
2 4/7 x 3 1/9 x 1 1/4 = 10
The minimum length of DF is 39.
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.
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.
Because of similar triangles, the two ships will each reach their destination at the same time.
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.
The longest possible streak is 78.
The first such streak starts with 9,999,999,961 and continues until 10,000,000,038.
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).
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
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.
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.
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.
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.
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.
