Puzzles
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.
4 December
If \(n\) is 1, 2, 4, or 6 then \((n!-3)/(n-3)\) is an integer. The largest of these numbers is 6.
What is the largest possible value of \(n\) for which \((n!-123)/(n-123)\) is an integer?
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We can split this fraction into:
$$\frac{n!}{n-123}-\frac{123}{n-123}$$
As long as \(n\) is larger than 123, \(n-123\) is positive and less than \(n\). This means that \(n!/(n-123)\) is an integer, as \(n-123\) will be one of the numbers multiplied together in \(n!\).
(\((124!-123)/(124-123)\) is an integer, so the answer is at least 124 and \(n!/(2-123)\) is an integer when \(n\) is the answer.)
We now need to answer the simpler question: What is the largest value of \(n\) for which \(123/(n-123)\) is an integer? This will be when \(n-123\) is equal to 123, and so \(n\) is 246.
3 December
190 is the smallest multiple of 10 whose digits add up to 10.
What is the smallest multiple of 15 whose digits add up to 15?
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