Puzzles
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?
2 December
Holly adds up the first six even numbers, then adds on half of the next even number. Her total is 49.
Next, Holly adds up the first \(n\) even numbers then adds on half of the next even number. This time, her total is 465124. What is \(n\)?
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If we add up the first \(n\) even numbers then add on half of the next even number, we get \((n+1)^2\). This means that Holly added up the first \(\sqrt{465124}-1\) or 681 even numbers.
Extension
Can you show why adding up the first \(n\) even numbers and half of the next even number gives \((n+1)^2\)?
1 December
Each interior angle of a regular triangle is 60°.
Each interior angle of a different regular polygon is 178°. How many sides does this polygon have?
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The exterior angles of any polygon add up to 360°.
The interior angles of our polygon are 178°, and so the exterior angles are 2°. There must be 180 sides (and so also 180 angles) on the polygon, as 180×2=360.