Puzzles
8 December
Carol uses the digits from 0 to 9 (inclusive) exactly once each to write five 2-digit even numbers, then finds their sum. What is the largest number she could have obtained?
6 December
Noel's grandchildren were in born in November in consecutive years. Each year for Christmas, Noel gives each of his grandchildren their age in pounds.
Last year, Noel gave his grandchildren a total of £208. How much will he give them in total this year?
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The only way to make 208 by adding consecutive numbers is
10+11+12+13+14+15+16+17+18+19+20+21+22. Therefore next year Noel will give his grandchildren a total of £221.
4 December
There are 5 ways to tile a 3×2 rectangle with 2×2 squares and 2×1 dominos.
Today's number is the number of ways to tile a 9×2 rectangle with 2×2 squares and 2×1 dominos.
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Let \(a_n\) be the number of ways to tile an \(n\times2\) rectangle.
There is one way to tile a 1×2 rectangle; and there are three ways to tile a 2×2 rectangle. Therefore \(a_1=1\) and \(a_2=3\).
In a \(n\times2\) rectangle, the rightmost column will either contain a vertical 2×1 domino, two horizontal 2×1 dominoes, or a 2×2 square.
Therefore \(a_n=a_{n-1}+2a_{n-2}\).
Using this, we find that there are 341 ways to tile a 9×2 rectangle.
3 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 largest number you can make with the digits in the red boxes.
| + | | + | | = 21 |
| + | | × | | × | |
| + | | + | | = 10 |
| + | | ÷ | | × | |
| + | | + | | = 14 |
=
21 | | =
10 | | =
14 | |
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| 8 | + | 6 | + | 7 | = 21 |
| + | | × | | × | |
| 4 | + | 5 | + | 1 | = 10 |
| + | | ÷ | | × | |
| 9 | + | 3 | + | 2 | = 14 |
=
21 | | =
10 | | =
14 | |
The largest number you can make with the digits in the red boxes is 432.
2 December
You have 15 sticks of length 1cm, 2cm, ..., 15cm (one of each length). How many triangles can you make by picking three sticks and joining their ends?
Note: Three sticks (eg 1, 2 and 3) lying on top of each other does not count as a triangle.
Note: Rotations and reflections are counted as the same triangle.
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In order to make a valid triangle, the sum of the two shorter sides must be larger than the longest side.
There is
1 triangle with longest side 4cm (2,3,4);
2 with longest side 5cm (2,4,5 and 3,4,5);
4 with longest side 6cm (2,5,6; 3,5,6; 4,5,6 and 3,4,6);
6 with longest side 7cm (2,6,7; 3,6,7; 4,6,7; 5,6,7; 3,5,7 and 4,5,7);
9 with longest side 8cm;
12 with longest side 9cm;
16 with longest side 10cm;
20 with longest side 11cm;
25 with longest side 12cm;
30 with longest side 13cm;
36 with longest side 14cm; and
42 with longest side 15cm.
In total this makes 203 triangles.
1 December
If you write out the numbers from 1 to 1000 (inclusive), how many times will you write the digit 1?
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100 numbers will have 1 in the units column. Another 100 with have a 1 in the tens column; and other 100 will have a 1 in the hundreds column.
Finally, 1000 has a 1 in the thousands column.
In total, this makes 301 copies of the digit 1.
Not Roman numerals
The letters \(I\), \(V\) and \(X\) each represent a different digit from 1 to 9. If
$$VI\times X=VVV,$$
what are \(I\), \(V\) and \(X\)?
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For any digit \(V\), \(VVV\) is a multiple of 111, and \(111=37\times3\). 37 is prime, so \(VI\) must be a multiple of 37 (as \(X\) is less than 10 so cannot be a multiple of 37). Therefore \(VI\) is either 37 or 74.
If \(VI\) was 74, then \(VVV\) is 777. But then \(VI\times X\) is even and \(VVV\) is odd. This is impossible, so \(VI\) must the 37.
Now that we know that \(VI\) is 37, we see that
$$37\times X=333,$$
and so \(X=9\) and the final solution is
$$37\times9=333.$$
24 December
1,0,2,0,1,1
The sequence of six numbers above has two properties:
- Each number is either 0, 1 or 2.
- Each pair of consecutive numbers adds to (strictly) less than 3.
Today's number is the number of sequences of six numbers with these two properties
