Advent calendar 2020

Santa's house could be found by following the directions 3,7,1,1,4,3,3,4,9,3,3,9.

Either both pieces must be on the diagonal, or one pieces is in the lower right half and the other is in the reflected position in the upper right half.

There are \(\left(\begin{array}{c}29\\2\end{array}\right)=406\) ways to pick two squares on the diagonal. There are 406 squares below the diagonal.

Therefore there are 406+406 = 812 ways to arrange the pieces.

We're looking for a number \(n\) such that the digits of \(n\times48\) add up to \(n\). If we try numbers in order, we find that 9 works, so today's number is 432.

The largest number you can make with the digits in the red boxes is 984.

To make sequences of 4 numbers, we can insert 4s into three different places in the length 3 sequences given to obtain:

There are some possibilities missing: those containing \(i, 4, i+1\). These can be found by taking the sequences of length 2, picking a number \(i\), adding 1 to every number larger than \(i\), then replacing \(i\) with \(i\ 4\ i+1\).

This gives a total of 3×3+2×1=11 sequences for 4 numbers.

To make sequences with 5 numbers, we can insert 5s into four different places in the length 4 sequences. This gives 4×11=44 sequences.

The missing sequences can then be found by taking the sequences of length 3, then doing the same process as above:

There are 3×3=9 of these, giving 44+9 = 53 total sequences of length 5.

To make sequences with 6 numbers, we can insert 6s into five different places in the length 5 sequences. This gives 5×53=265 sequences. We can also make sequence by picking a number to replace in the length 4 sequences. This gives 4×11=44 more sequences. Therefore there are 265+44 = 309 sequences in total.

The sum of 13 consecutive integers is 13 times the middle number, so today's number is a multiple of 13. The sum of 12 consecutive integers is 6 times the sum of the two middle numbers, so today's number is also a multiple of 6. Therefore today's number is a multiple of 78.

78 and 156 do not work (the numbers would not all be strictly positive), so today's number is 234.

Each triangle must include one of the two corners at the base of the largest triangle (or both those corners).

To make triangles including the bottom left corner of the large triangle, we must pick two lines coming out of that corner, and one line coming out of the bottom right corner that is not the base of the triangle. There are 9 lines coming out of the corners, so the total number of triangles is \(\left(\begin{array}{c}9\\2\end{array}\right)\times\left(\begin{array}{c}8\\1\end{array}\right)=288\).

We now need to count the triangles that include the bottom right corner of the large triangle, but do not include both corners (as we've already counted thoses). There are \(\left(\begin{array}{c}8\\2\end{array}\right)\times\left(\begin{array}{c}8\\1\end{array}\right)=224\) ways to pick lines to make these triangles.

In total, this makes 512 triangles.

In each term, the powers of \(x\), \(y\) and \(z\) must add to 26: the terms will be of the form \(x^iy^jz^{26-i-j}\).

If \(i=0\), there are 27 choices for \(j\) (0 to 26). If \(i=1\), there are 26 choices for \(j\) (0 to 25). If \(i=2\), there are 25 choices for \(j\) (0 to 24). ... If \(i=26\), there is 1 choice for \(j\) (0).

Therefore the total number of terms is 27+26+25+...+1 = 378.

The product of the numbers in the red boxes is 189.

Today's number is 333.

SAM is 781.

There are 81 two-digit numbers with no duplicate digits: there are 9 choices of tens digit (1 to 9), and for each of these there are 9 remaining choices for the units digit (0 to 9 but not the number already used).

There are 648 three-digit numbers with no duplicate digits: there are 9 choices of hundreds digit (1 to 9), and for each of these there are 9 remaining choices for the tens digit (0 to 9 but not the number already used) and 8 choices for the units digit (0 to 9 but neight number already used).

648+81 = 729.

There are 9 gaps between the numbers 1 to 10. We need to pick 5 of these gaps to split the sequence. There are \(\left(\begin{array}{c}9\\5\end{array}\right)\) ways to do this: this is 126.

No matter the exact size of each square, the blue quadrilateral will always fill half the square, so the area of the square is 334.

Let's say there were originally \(t\) cards of each colour, and \(2a\) red cards and \(a\) black cards in pile A. After the move, there were \(t-2a+108\) red cards and \(t-a\) black cards in pile B. Two thirds of the card in pile B are red, so:

Therefore there were 108×2=216 cards.

Interestingly, it appears that this answer is independent of \(a\): any number of cards could be put in each pile and this situation would still work. However, only one situation could have happened unless you allow there to at some points have been a negative number of cards in each pile.

There are one three-digit even numbers whose digits add to 24: 798, 888 and 996. Of these, only 888 is a multiple of 24.

The product of the numbers in the red boxes is 144.

The Octingham resident's 11 is equal to our number 9. 9 squared is 81. 81 in base eight is 121. Interestingly, this is the same answer as "just" doing 11 squred in base ten.

Each number will appear 7 times: one time paired with the numbers six numbers 5 to 10, plus an extra appearance on the tile containing the same number twice. The total of all the numbers is therefore 7×(5+6+...+10)=7×45=315.

First, consider placing 5 tiles in a 1×9 grid. There is only one way to do this:

To get the number of ways of placing 5 tiles in a 1×10 grid, imagine adding an extra blank square to either the start or end of the grid or between two of the counters. There are 6 places this tile could be inserted leading to 6 arrangements of 5 tiles in a 1×10 grid.

For 5 tiles in a 2×10 grid, you can first pick the columns the tiles go in (as a tile being in a column means nothing can be placed the columns either side, the number of ways to pick columns is the same and the number of wats to arrange 5 tokens in a 1×10 grid). For each of these column choices, there are two locations for each tile (top or bottom). This leads to a total number of arrangements of 6×2⁵=192.

The totals that are equally likely add up to 7 times the number of dice.

This can be seen by using the fact that the opposite sides of a dice add up to 7: for each way of making a given total with \(n\) dice, there is a way of making \(7n\) minus that total by looking at the dice from below. Therefore \(T\) and \(7n-T\) are equally likely. (This also holds true (but is harder to explain) if you rearrange the faces of the dice so that the opposite faces no longer add to 7.)

Therefore today's number is \((521+200)/7\), which is 103.

If the final digit of the number is 0, then some carrying takes place when 1 is subtracted. Otherwise, no carrying happens.

If no carrying happens, call the three digits of today's number \(A\), \(B\), and \(C\). We know that \(A^3+B^3+C^3\) is one more than \(A^3 + B^3 + (C-1)^3\). This implies that \(C^3=(C-1)^3+1\), which is only possible if \(C\) is 1.

Therefore either the final digit of today's number is 0 or the final digt of one less that today's number is 0. In both cases, we need to find a number with the desired property whose final digit is 0: we are looking for digit \(A\) and \(B\) such that \(A^3+B^3\) is a multiple of 10.

Looking at all the cube numbers, there are a few combinations that add up to multiple of 10:

The only one of these that has the required property is 370. By checking 369 and finding it doesn't have the property, we see that the two numbers must be 370 and 371.

The largest number you can make with the digits in the red boxes is 321.

By drawing an appropriate diagram, it can be seen that the small square has half the area of the large square.

Therefore the area of the large square is 124.

There are 3 one-digit numbers using these digits (1, 3 and 5).

To make two-digit odd numbers, there are 3 choices for the units digit, and 4 remaining choices for the tens digit.

To make three-digit odd numbers, there are 3 choices for the units digit, 4 remaining choices for the tens digit, and 3 remaining choices for the hundreds digit.

To make four-digit odd numbers, there are 3 choices for the units digit, 4 remaining choices for the tens digit, 3 remaining choices for the hundreds digit, and 2 remaining choices for the thousands digit.

To make five-digit odd numbers, there are 3 choices for the units digit, 4 remaining choices for the tens digit, 3 remaining choices for the hundreds digit, 2 remaining choices for the thousands digit and 1 remaining choice for the ten-thousands digit.

In total, this gives 3 + 3×4× + 3×4×3 + 3×4×3×2 + 3×4×3×2×1 = 195 odd numbers.