Puzzles
6 December
There are 21 three-digit integers whose digits are all non-zero and whose digits add up to 8.
How many positive integers are there whose digits are all non-zero and whose digits add up to 8?
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There is 1 number whose digits are all non-zero and whose digits add up to 1 (1).
There are 2 numbers whose digits are all non-zero and whose digits add up to 2 (2, and 11).
There are 4 numbers whose digits are all non-zero and whose digits add up to 3 (3, 12, 21, and 111).
There are 8 numbers whose digits are all non-zero and whose digits add up to 4 (4, 13, 31, 112, 121, 211, and 1111).
The amount of numbers is doubling each time. You can justify this by noticing that every number whose digits are all non-zero and whose digits add to \(n+1\)
can be made from a number adding to \(n\) by either adding 1 to the final digit or appending a 1 onto the end of the number.
Therefore, there are 128 numbers whose digits are all non-zero and whose digits add up to 8.
Extension
There is 1 number whose digits are all non-zero and whose digits add up to 1.
There are 2 numbers whose digits are all non-zero and whose digits add up to 2.
There are 4 numbers whose digits are all non-zero and whose digits add up to 3.
There are 8 numbers whose digits are all non-zero and whose digits add up to 4.
There are 16 numbers whose digits are all non-zero and whose digits add up to 5.
There are 32 numbers whose digits are all non-zero and whose digits add up to 6.
There are 64 numbers whose digits are all non-zero and whose digits add up to 7.
There are 128 numbers whose digits are all non-zero and whose digits add up to 8.
There are 256 numbers whose digits are all non-zero and whose digits add up to 9.
Are there 512 numbers whose digits are all non-zero and whose digits add up to 10?
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 |
| + | | + | | + | |
| × | | ÷ | | = 14 |
| – | | – | | – | |
| × | | ÷ | | = 27 |
=
9 | | =
5 | | =
5 | |
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| 5 | × | 6 | ÷ | 2 | = 15 |
| + | | + | | + | |
| 7 | × | 8 | ÷ | 4 | = 14 |
| – | | – | | – | |
| 3 | × | 9 | ÷ | 1 | = 27 |
=
9 | | =
5 | | =
5 | |
The product of the numbers in the red boxes is 315.
3 December
Write the numbers 1 to 81 in a grid like this:
$$
\begin{array}{cccc}
1&2&3&\cdots&9\\
10&11&12&\cdots&18\\
19&20&21&\cdots&27\\
\vdots&\vdots&\vdots&\ddots&\vdots\\
73&74&75&\cdots&81
\end{array}
$$
Pick 9 numbers so that you have exactly one number in each row and one number in each column,
and find their sum.
What is the largest value you can get?
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If you add these red numbers to the start of each row and column, then every number in the grid is the sum of the red numbers in its row and column
\begin{array}{ccccc}
&{\color{red}1}&{\color{red}2}&{\color{red}3}&\cdots&{\color{red}9}\\
{\color{red}0}&0+1&0+2&0+3&\cdots&0+9\\
{\color{red}9}&9+1&9+2&9+3&\cdots&9+9\\
{\color{red}18}&18+1&18+2&18+3&\cdots&18+9\\
\vdots&\vdots&\vdots&\vdots&\ddots&\vdots\\
{\color{red}72}&72+1&72+2&72+3&\cdots&72+9
\end{array}
However you pick one number from each row and column, you will always end up with the total of all the red numbers. This is 369.
24 December
The digital product of a number is computed by multiplying together all of its digits.
For example, the digital product of 1522 is 20.
How many 12-digit numbers are there whose digital product is 20?
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20 can be written as the product of one-digit numbers greater than 1 in the following ways:
Our 12-digit numbers must contain one of these plus lots of 1s.
In the first case, there are 12×11=132 different possible positions for the 4 and the 5.
In the second case, there are 12×11×10×½=660 different possible positions for the 2s and the 5.
In total, this makes 132+660=792 numbers.
22 December
There are 12 ways of placing 2 tokens on a 2×4 grid so that no two tokens are next to each other horizontally, vertically or diagonally:
Today's number is the number of ways of placing 2 tokens on a 2×21 grid so that no two tokens are next to each other horizontally, vertically or diagonally.
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The two tokens must be in two non-adjacent columns. There are ½×21×20 ways of picking two different columns. 20 of these ways will give two adjacent columns,
so there are ½×21×20–20=190 ways to pick the columns.
Once the columns are picked there are four choices for the rows to place the tokens in (up and up, up and down, down and up, down and down). 4×190=760.
21 December
Arrange the digits 1–9 (using each digit exactly once) so that the three digit number in:
the middle row is a prime number;
the bottom row is a square number;
the left column is a cube number;
the middle column is an odd number;
the right column is a multiple of 11.
The 3-digit number in the first row is today's number.
| | | today's number |
| | | prime |
| | | square |
| cube | odd | multiple of 11 |
18 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.
| + | | + | | = 11 |
| + | | × | | × | |
| + | | + | | = 17 |
| × | | - | | + | |
| + | | + | | = 17 |
=
11 | | =
17 | | =
17 | |
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| 6 | + | 3 | + | 2 | = 11 |
| + | | × | | × | |
| 5 | + | 8 | + | 4 | = 17 |
| × | | - | | + | |
| 1 | + | 7 | + | 9 | = 17 |
=
11 | | =
17 | | =
17 | |
The product of the numbers in the red boxes is 432.
17 December
The digital product of a number is computed by multiplying together all of its digits.
For example, the digital product of 6273 is 252.
Today's number is the smallest number whose digital product is 252.
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252 can be written as the product of one-digit numbers in the following ways:
- 2×2×3×3×7
- 2×2×9×7
- 2×3×6×7
- 3×3×4×7
- 6×6×7
- 4×9×7
Therfore, the smallest number whose digital product is 252 is 479.
