Advent calendar 2019

Santa was at 36.11.19.79.79.43. Rudolph was at 16.64.16.16.16.64. Blitzen was at 23.12.23.11.23.11.

The digits of the number must be 1, 2 and 3, so the largest number is 321.

The two numbers are 243 and 244.

The smallest number you can make with the digits in the red boxes is 123.

Because 13 and 11 are prime and larger than 7, no-one can have taken them, as their is no way the other person's product could then be the same. Once 13 and 11 are discarded, the prime factorisation of the product of all the other cards is \(2^{11}\times3^5\times5^2\times7^2\). In order to be shared to give the same product, the powers must be even, and so the other card not taken must be 6 (as 2 and 3 are the numbers with odd powers).

Therefore the product of the three cards is 858.

The diagonal of the blue square is \(\sqrt{28}\). The radius \(R\) of the large circle satisfies \(R^2+R^2=(R+\sqrt{28})^2\). Solving this, we find that \(R=\frac{\sqrt{28}}{\sqrt2-1}=\sqrt{28}(\sqrt2+1)\).

The radius \(r\) of the small circles satisfies \(r+r\sqrt2=R\), and so \(r=\frac{\sqrt{28}(\sqrt2+1)}{\sqrt2+1}=\sqrt{28}\).

The area of the square is \(4r^2=4\left(\sqrt{28}\right)^2=4\times28\). This is 112.

To win this game, you need to always end your turn on a multiple of 901, as whatever you opponent does, you can always get back to a multiple of 901.

Therefore, you should take £791 on your first turn to get to a multiple of 901.

For the first and last digits of the result to be different, there must be a 1 carried from the middle column. Therefore the middle digit of Eve's number is 9. Eve's number could be 292 or 193, but it cannot be the first as reversing this is not larger than her original number, so here number is 193.

30030 is 2×3×5×7×11×13, so the answer is the number of ways of sharing six elements into unlabelled boxes (A000110).

The answer is 203.

There are 1 time starting 00; 2 times starting 01; 3 times starting 02; 4 times starting 03; 5 times starting 04; 6 times starting 05; 6 times starting 06; 6 times starting 07; 6 times starting 08; 6 times starting 09; 2 times starting 10; 3 times starting 11; 4 times starting 12; 5 times starting 13; 6 times starting 14; 6 times starting 15; 6 times starting 16; 6 times starting 17; 6 times starting 18; 5 times starting 19; 3 times starting 20; 4 times starting 21; 5 times starting 22; and 6 times starting 23.

Therefore there are 112 ways in total.

The Rum party needs a minimum of 50 votes to win a consituency, so they could have at most 32175÷50=643 MAPs.

The product of the red digits is 192.

If we plot the two quadratics, we see that they meet at \(x=2\) where they both have gradient 4.

The line \(f(x)\) must therefore pass through the point \((2,4)\) and have gradient 4, so its equation is \(f(x)=4x-4\).

This means that \(f(65)\) is 256.

Carol will get 270 whatever she does.

The sum of the coefficients can be found by substituting \(x=1\) into the expression. Hence the sum of the coefficients in the expansion of \((2x+1)^5\) is \(3^5\), or 243.

The only way to make 208 by adding consecutive numbers is 10+11+12+13+14+15+16+17+18+19+20+21+22. Therefore next year Noel will give his grandchildren a total of £221.

Pick one of the 28 points and imagine a line to the point opposite it. There will be 13 points on each side of this line. Picking one of these 13 points and then picking the corresponding point on the other side of the line gives an isosceles triangle. Therefore there are 13 isosceles triangles with the first chosen point as the point where the two equal sides meet.

There were 28 choices for the first point, and so the total number of isosceles triangles will be 13×28=364.

Let \(a_n\) be the number of ways to tile an \(n\times2\) rectangle.

There is one way to tile a 1×2 rectangle; and there are three ways to tile a 2×2 rectangle. Therefore \(a_1=1\) and \(a_2=3\).

In a \(n\times2\) rectangle, the rightmost column will either contain a vertical 2×1 domino, two horizontal 2×1 dominoes, or a 2×2 square. Therefore \(a_n=a_{n-1}+2a_{n-2}\).

Using this, we find that there are 341 ways to tile a 9×2 rectangle.

The largest number you can make with the digits in the red boxes is 432.

In order to make a valid triangle, the sum of the two shorter sides must be larger than the longest side. There is 1 triangle with longest side 4cm (2,3,4); 2 with longest side 5cm (2,4,5 and 3,4,5); 4 with longest side 6cm (2,5,6; 3,5,6; 4,5,6 and 3,4,6); 6 with longest side 7cm (2,6,7; 3,6,7; 4,6,7; 5,6,7; 3,5,7 and 4,5,7); 9 with longest side 8cm; 12 with longest side 9cm; 16 with longest side 10cm; 20 with longest side 11cm; 25 with longest side 12cm; 30 with longest side 13cm; 36 with longest side 14cm; and 42 with longest side 15cm.

In total this makes 203 triangles.

100 numbers will have 1 in the units column. Another 100 with have a 1 in the tens column; and other 100 will have a 1 in the hundreds column. Finally, 1000 has a 1 in the thousands column.

In total, this makes 301 copies of the digit 1.