Puzzles
18 December
Noel writes the integers from 1 to 1000 in a large triangle like this:
| | | | 1 | | | |
| | | 2 | 3 | 4 | | |
| | 5 | 6 | 7 | 8 | 9 | |
| 10 | 11 | 12 | 13 | ... | | |
The number 12 is directly below the number 6. Which number is directly below the number 133?
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The rightmost entry in the \(n\)th row is \(n^2\). A number below a number in the \(n\)th row is \(2n\) larger than the number in the \(n\)th row.
The smallest square number larger than 133 is 144, so 133 is in the 12th row. This means that the number under it will be 24 greater than it: 133+24=157.
17 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 product of the numbers in the red boxes.
| + | | + | | = 10 |
| + | | × | | × | |
| + | | + | | = 12 |
| + | | – | | + | |
| + | | + | | = 23 |
=
10 | | =
12 | | =
23 | |
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| 3 | + | 5 | + | 2 | = 10 |
| + | | × | | × | |
| 1 | + | 4 | + | 7 | = 12 |
| + | | – | | + | |
| 6 | + | 8 | + | 9 | = 23 |
=
10 | | =
12 | | =
23 | |
The product of the numbers in the red boxes is 144.
16 December
Noel writes the integers from 1 to 1000 in a large triangle like this:
| | | | 1 | | | |
| | | 2 | 3 | 4 | | |
| | 5 | 6 | 7 | 8 | 9 | |
| 10 | 11 | 12 | 13 | ... | | |
The rightmost number in the row containing the number 6 is 9.
What is the rightmost number in the row containing the number 300?
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The rightmost numbers in each row are the square numbers. The smallest square number larger than 300 is 324.
15 December
There are 3 even numbers between 3 and 9.
What is the only odd number \(n\) such that there are \(n\) even numbers
between \(n\) and 729?
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There are \((729-n)/2\) even numbers between \(n\) and 729, and so we want to solve \((729-n)/2=n\). The solution of this is 243.
Extension
For which odd numbers \(N\) does there exist an odd number \(n\) such that there are \(n\) even numbers between \(n\) and \(N\)?
14 December
Holly draws a line of connected regular pentagons like this:
She continues the pattern until she has drawn 204 pentagons. The perimeter of each pentagon is 5.
What is the perimeter of her line of pentagons?
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The two pentagons at the end of the pattern each contribute 4 units to the permimeter. The remaining 202 pentagons each contribute 3 units to the perimeter.
This gives a total perimeter of 2×4+202×3=614.
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.
|
|
| Across |
| 1 | A cube number. | (3) |
| 4 | A square number. | (3) |
| 5 | A multiple of 13. | (3) |
|
Down
(in a random order)
|
- Two times 1A.
- A square number.
- Today's number.
|
|
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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
\(\{1, 2, 3, ..., 12\}\)?
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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.
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