Puzzles
13 December
Today's number is given in this crossnumber. The across clues are given as normal, but the down clues are given in a random order: you must work out
which clue goes with each down entry and solve the crossnumber to find today's number.
No number in the completed grid starts with 0.
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| Across |
| 1 | A cube number. | (3) |
| 4 | A square number. | (3) |
| 5 | A multiple of 13. | (3) |
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Down
(in a random order)
|
- Today's number.
- A square number.
- Two times 1A.
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|
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1D is "a square number"; 2D is "two times 1A"; 3D is "today's number": today's number is 564.
12 December
The determinant of the 2 by 2 matrix \(\begin{pmatrix}a&b\\c&d\end{pmatrix}\) is \(ad-bc\).
If a 2 by 2 matrix's entries are all in the set \(\{1, 2, 3\}\), the largest
possible deteminant of this matrix is 8.
What is the largest possible determinant of a 2 by 2 matrix whose entries are all in the set
\(\{1, 2, 3, ..., 12\}\)?
Show answer & extension
Hide answer & extension
The largest deteminant will be made by making \(a\) and \(d\) as large as possible (ie 12) and \(b\) and \(c) as small as possible (ie 1). This gives a
determinant of 143.
Extension
The determinant of the 3 by 3 matrix \(\begin{pmatrix}a&b&c\\d&e&f\\g&h&i\end{pmatrix}\) is \(a(ei-fh)-b(di-fg)+c(dh-eg)\).
If a 3 by 3 matrix's entries are all in the set \(\{1, 2, 3\}\), the largest possible deteminant of this matrix is 28.
What is the largest possible determinant of a 3 by 3 matrix whose entries are all in the set
\(\{1, 2, 3, ..., 12\}\)?
11 December
There are five 3-digit numbers whose digits are all either 1 or 2 and who do not contain
two 2s in a row: 111, 112, 121, 211, and 212.
How many 14-digit numbers are there whose digits are all either 1 or 2 and who do not contain
two 2s in a row?
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There are two 1-digit numbers whose digits are all either 1 or 2 and who do not contain two 2s in a row (1, and 2).
There are three 2-digit numbers whose digits are all either 1 or 2 and who do not contain two 2s in a row (11, 12, and 21).
There are five 3-digit numbers whose digits are all either 1 or 2 and who do not contain two 2s in a row (111, 112, 121, 211, and 212).
There are eight 4-digit numbers whose digits are all either 1 or 2 and who do not contain two 2s in a row (1111, 1112, 1121, 1211, 1212, 2111, 2112, and 2121).
(It looks like these are the Fibonacci numbers.)
All these numbers either end in a 1, or end in 12. Therefore the \((n+1)\)-digit number can be made by appending a 1 to the end of the \(n\)-digit numbers
or appending a 12 to the end of the \((n-1)\)-digit numbers: so the next term is always the sum of the previous two terms.
Continuing the pattern, we see that there are 987 14-digit numbers are there whose digits are all either 1 or 2 and who do not contain
two 2s in a row.
10 December
A line is tangent to a curve if the line touches the curve at exactly one point.
The line \(y=-160\,000\) is tangent to the parabola \(y=x^2-ax\). What is \(a\)?
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A line \(y=\text{constant}\) that is tangent to a parabola must be a tangent to the minimum (or maximum) point.
\(y=x^2-ax\) can be rewritten as \(y=(x-a/2)^2-a^2/4\). The minimum of this will be at \(y=-a^2/4\).
If \(-a^2/4=-160\,000\), then \(a\) is 800.
9 December
Put the digits 1 to 9 (using each digit exactly once) in the boxes so that the sums are correct. The sums should be read left to right and top to bottom ignoring the usual order of operations. For example, 4+3×2 is 14, not 10.
Today's number is the largest number you can make with the digits in the red boxes.
| + | | + | | = 20 |
| + | | + | | ÷ | |
| + | | – | | = 0 |
| + | | – | | × | |
| ÷ | | × | | = 12 |
=
22 | | =
6 | | =
2 | |
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| 9 | + | 7 | + | 4 | = 20 |
| + | | + | | ÷ | |
| 5 | + | 1 | – | 6 | = 0 |
| + | | – | | × | |
| 8 | ÷ | 2 | × | 3 | = 12 |
=
22 | | =
6 | | =
2 | |
The largest number you can make with the digits in the red boxes is 532.
8 December
The equation \(x^5 - 7x^4 - 27x^3 + 175x^2 + 218x = 840\) has five real solutions. What is the product of all these solutions?
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The product of all the solutions is 840.
Expanding \((x-\alpha)(x-\beta)(x-\gamma)(x-\delta)(x-\epsilon)=0\) gives \(x^5 + \dots = \alpha\beta\gamma\delta\epsilon\). As long as the
coefficient of \(x^5\) is 1, the constant term when written like this will be the product of all the solutions (for this to always work, you will need to include repeated solutions and complex solutions.)
Extension
The equation \(x^8 - 19x^7 + 126x^6 - 294x^5 - 231x^4 + 1869x^3 - 1576x^2 - 1556x + 1680=0\) has eight real solutions. What is the product of all these solutions?
7 December
What is the area of the largest triangle that fits inside a regular hexagon with area 952?
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The vertices of this triangle will be every other vertex of the hexagon: and other triangle can be made larger by moving one of its vertices closer to a vertex of the hexagon.
The area of this triangle is 476.
6 December
There are 21 three-digit integers whose digits are all non-zero and whose digits add up to 8.
How many positive integers are there whose digits are all non-zero and whose digits add up to 8?
Show answer & extension
Hide answer & extension
There is 1 number whose digits are all non-zero and whose digits add up to 1 (1).
There are 2 numbers whose digits are all non-zero and whose digits add up to 2 (2, and 11).
There are 4 numbers whose digits are all non-zero and whose digits add up to 3 (3, 12, 21, and 111).
There are 8 numbers whose digits are all non-zero and whose digits add up to 4 (4, 13, 31, 112, 121, 211, and 1111).
The amount of numbers is doubling each time. You can justify this by noticing that every number whose digits are all non-zero and whose digits add to \(n+1\)
can be made from a number adding to \(n\) by either adding 1 to the final digit or appending a 1 onto the end of the number.
Therefore, there are 128 numbers whose digits are all non-zero and whose digits add up to 8.
Extension
There is 1 number whose digits are all non-zero and whose digits add up to 1.
There are 2 numbers whose digits are all non-zero and whose digits add up to 2.
There are 4 numbers whose digits are all non-zero and whose digits add up to 3.
There are 8 numbers whose digits are all non-zero and whose digits add up to 4.
There are 16 numbers whose digits are all non-zero and whose digits add up to 5.
There are 32 numbers whose digits are all non-zero and whose digits add up to 6.
There are 64 numbers whose digits are all non-zero and whose digits add up to 7.
There are 128 numbers whose digits are all non-zero and whose digits add up to 8.
There are 256 numbers whose digits are all non-zero and whose digits add up to 9.
Are there 512 numbers whose digits are all non-zero and whose digits add up to 10?
