Puzzles
19 December
The diagram to the right shows a triangle. Two of the sides of the triangle have been split into three pieces, with lines drawn from the opposite vertex. In total, the diagram now contains 27 triangles of any size.
Another triangle has two of its sides split into eight pieces, with lines drawn from the opposite vertex. How many triangles (of any size) would this create?
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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.
18 December
The expansion of \((x+y+z)^3\) is
$$x^3 + y^3 + z^3 + 3x^2y + 3x^2z + 3xy^2 + 3y^2z + 3xz^2 + 3yz^2 + 6xyz.$$
This has 10 terms.
Today's number is the number of terms in the expansion of \((x+y+z)^{26}\).
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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.
17 December
Put the digits 1 to 9 (using each digit exactly once) in the boxes so that the sums and product are correct.
Today's number is the product of the numbers in the red boxes.
| + | | + | | = 16 |
| + | | + | | + | |
| + | | + | | = 8 |
| + | | + | | + | |
| × | | × | | = 288 |
=
11 | | =
14 | | =
20 | |
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| 6 | + | 3 | + | 7 | = 16 |
| + | | + | | + | |
| 1 | + | 2 | + | 5 | = 8 |
| + | | + | | + | |
| 4 | × | 9 | × | 8 | = 288 |
=
11 | | =
14 | | =
20 | |
The product of the numbers in the red boxes is 189.
16 December
Solve the crossnumber to find today's number. No number starts with 0.
|
|
| Across |
| 1 | 3 more than a multiple of 110. | (3) |
| 4 | Today's number. | (3) |
| 5 | 30 more than a multiple of 101. | (3) |
| Down |
| 1 | 37 times the sum of 1A's digits. | (3) |
| 2 | 3 times a factor of 5A. | (3) |
| 3 | 3 less than a multiple of 112. | (3) |
|
15 December
When talking to someone about this Advent calendar, you told them that the combination of XMAS and MATHS is GREAT.
They were American, so asked you if the combination of XMAS and MATH is great; you said SURE. You asked them their name; they said SAM.
Each of the letters E, X, M, A, T, H, S, R, U, and G stands for a different digit 0 to 9. The following sums are correct:
Today's number is SAM. To help you get started, the letter T represents 4.
14 December
The numbers 33, 404 and 311 contain duplicate digits. The numbers 120, 15 and 312 do not.
How many numbers between 10 and 999 (inclusive) contain no duplicate digits?
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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.
13 December
There are 6 ways to split the sequence of the numbers 1 to 5 into three shorter sequences:
- 1 and 2 and 3, 4, 5
- 1 and 2, 3 and 4, 5
- 1 and 2, 3, 4 and 5
- 1, 2 and 3 and 4, 5
- 1, 2 and 3, 4 and 5
- 1, 2, 3 and 4 and 5
Today's number is the number of ways to split the sequence of the numbers 1 to 10 into five shorter sequences.
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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.
12 December
The diagram to the left shows a large black square. Inside this square, two red squares have been drawn.
(The sides of the red squares are parallel to the sides of the black square; each red square shares a vertex with the black square; and the two red squares share a vertex.)
A blue quadrilateral has then been drawn with vertices at two corners of the black square and the centres of the red squares.
The area of the blue quadrilateral is 167. What is the area of the black square?
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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.
![](javascript:showlimage("advent2020triangles.png"))
![](javascript:showlimage("advent2020quads.png"))