Puzzles
12 December
What is the smallest value of \(n\) such that
$$\frac{500!\times499!\times498!\times\dots\times1!}{n!}$$
is a square number?
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Consider the first two terms in the product:
$$
500!\times499!
=500\times499!\times499!$$
$$= 500\times(499!)^2.$$
Doing similar steps with each pair of terms in the product, we see that:
$$
500!\times499!\times498!\times\dots\times1!
=
500\times498\times\dots\times2\times(499!\times497!\times\dots\times1!)^2
$$
$$
=
(2\times250)\times(2\times249)\times\dots\times(2\times1)\times(499!\times497!\times\dots\times1!)^2
$$
$$
=
2^{250}\times250!\times(499!\times497!\times\dots\times1!)^2
$$
If we divide this by \(250!\), we are left with a square number, and so \(n\) is 250
11 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.
| + | | + | | = 15 |
| + | | + | | ÷ | |
| + | | – | | = 10 |
| + | | – | | × | |
| ÷ | | × | | = 3 |
=
16 | | =
1 | | =
30 | |
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| 3 | + | 7 | + | 5 | = 15 |
| + | | + | | ÷ | |
| 9 | + | 2 | – | 1 | = 10 |
| + | | – | | × | |
| 4 | ÷ | 8 | × | 6 | = 3 |
=
16 | | =
1 | | =
30 | |
The product of the numbers in the red boxes is 120.
10 December
How many integers are there between 100 and 1000 whose digits add up to an even number?
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Between 100 and 109 (inclusive), there are 5 integers whose digits add up to an even number, and 5 whose digits add up to an odd number.
Between 110 and 119 (inclusive), there are 5 integers whose digits add up to an even number, and 5 whose digits add up to an odd number...
In general, between \(10n\) and \(10n+9\) (inclusive), there are 5 integers whose digits add up to an even number, and 5 whose digits add up to an odd number.
The integers from 100 to 999 (inclusive) can be split into 45 sets of integers from \(10n\) to \(10n+9\) (and the digits of 1000 don't add to an even number), so there are
450 integers between 100 and 1000 whose digits add up to an even number.
9 December
The diagram below shows a rectangle. Two of its sides have been coloured blue. A red line has been drawn from two of its vertices to the midpoint of a side.
The total length of the blue lines is 50cm. The total length of the red lines is also 50cm. What is the area of the rectangle (in cm2)?
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Let \(a\) be the height of the rectangle. As the total of the blue lines is 50, the width of the rectangle is \(50-a\).
As the total of the red lines is 50, each red line segment is 25.
Using Pythagoras's theorem in one of the right-angled triangles, we see that:
$$a^2 + \left(\frac{50-a}{2}\right)^2 = 25^2$$
$$4a^2 + (50-a)^2 = 50^2$$
$$4a^2 + 50^2 - 100a + a^2 = 50^2$$
$$a(5a - 100) = 0$$
\(a\) is not zero, and so \(a=20\). This means that the area of the rectangle is 20×30=600.
8 December
Noel writes the numbers 1 to 17 in a row. Underneath, he writes the same list without the first and last numbers, then continues this until he writes a row containing just one number:
| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 |
| | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | |
| | | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | | |
| etc. |
What is the sum of all the numbers that Noel has written?
7 December
There are 8 sets (including the empty set) that contain numbers from 1 to 4 that don't include any consecutive integers:
\(\{\}\), \(\{1\}\), \(\{2\}\), \(\{3\}\), \(\{4\}\), \(\{1,3\}\), \(\{1,4\}\), \(\{2, 4\}\)
How many sets (including the empty set) are there that contain numbers from 1 to 14 that don't include any consecutive integers?
6 December
There are 5 ways to tile a 4×2 rectangle with 2×1 pieces:
How many ways are there to tile a 12×2 rectangle with 2×1 pieces?
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Let \(a_n\) be the number of ways to tile a \(n\times2\) rectangle.
It is easy to check that \(a_1=1\) (ie there is 1 way to tile a 1×2 rectangle) and \(a_2=2\) (ie there are 2 ways to tile a 2×2 rectangle).
For an \(n\times2\) rectangle, from the left the tiling either starts with a vertical tile, or a pair of horizontal tiles.
If it starts with a vertical tile, then there are \(a_{n-1}\) ways to tile the remaining \((n-1)\times2\) rectangle.
If it starts with a pair of horizontal tile2, then there are \(a_{n-2}\) ways to tile the remaining \((n-2)\times2\) rectangle.
Therefore, \(a_n=a_{n-1}+a_{n-2}\).
(And so the number of ways to tile a \(n\times2\) rectangle is the \((n+1)\)th Fibonacci number.)
Therefore, the number of ways to tile a 12×2 rectangle is 233.
5 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.
| + | | + | | = 15 |
| + | | – | | + | |
| + | | + | | = 15 |
| + | | × | | ÷ | |
| + | | + | | = 15 |
=
15 | | =
15 | | =
15 | |
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| 2 | + | 6 | + | 7 | = 15 |
| + | | – | | + | |
| 4 | + | 3 | + | 8 | = 15 |
| + | | × | | ÷ | |
| 9 | + | 5 | + | 1 | = 15 |
=
15 | | =
15 | | =
15 | |
The product of the numbers in the red boxes is 378.
![](javascript:showlimage("advent2023-9dec.png"))
![](javascript:showlimage("advent2023-6dec.png"))