Puzzles
21 December
In the annual tournament of Christmas puzzles, each player must play one puzzle match
against each other player. Last year there were four entrants into the tournament (A, B, C, and D),
and so 6 matches were played: A vs B, C vs D, A vs D, A vs C, D vs B, and finally B vs C.
This year, the tournament has grown in popularity and 22 players have entered. How many
matches will be played this year?
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Each player must play 21 other players: this makes 22×21=462 matches. But this includes (eg) A vs B and B vs A as two different matches: every match is
included twice, so the number of matches if 462/2=231.
19 December
120 is the smallest number with exactly 16 factors (including 1 and 120 itself).
What is the second smallest number with exactly 16 factors (including 1 and the number itself)?
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If \(p_1^{a_1}\times p_2^{a_2}\times\dots\times p_n^{a_n}\) is the prime factorisation of a number, then the number has \((a_1+1)(a_2+1)\dots(a_n+1)\) factors.
The prime factorisation of 120 is \(2^3\times3\times5\). The next smallest number with 16 factors must be one of:
- \(2\times3^3\times5=270\)
- \(2^3\times3\times7=168\)
- \(2^3\times3^3=216\)
The smallest of these is 168.
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 \(N\)?
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.
