Puzzles
9 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.
| + | | + | | = 20 |
| + | | + | | ÷ | |
| + | | – | | = 0 |
| + | | – | | × | |
| ÷ | | × | | = 12 |
=
22 | | =
6 | | =
2 | |
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| 9 | + | 7 | + | 4 | = 20 |
| + | | + | | ÷ | |
| 5 | + | 1 | – | 6 | = 0 |
| + | | – | | × | |
| 8 | ÷ | 2 | × | 3 | = 12 |
=
22 | | =
6 | | =
2 | |
The largest number you can make with the digits in the red boxes is 532.
8 December
The equation \(x^5 - 7x^4 - 27x^3 + 175x^2 + 218x = 840\) has five real solutions. What is the product of all these solutions?
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The product of all the solutions is 840.
Expanding \((x-\alpha)(x-\beta)(x-\gamma)(x-\delta)(x-\epsilon)=0\) gives \(x^5 + \dots = \alpha\beta\gamma\delta\epsilon\). As long as the
coefficient of \(x^5\) is 1, the constant term when written like this will be the product of all the solutions (for this to always work, you will need to include repeated solutions and complex solutions.)
Extension
The equation \(x^8 - 19x^7 + 126x^6 - 294x^5 - 231x^4 + 1869x^3 - 1576x^2 - 1556x + 1680=0\) has eight real solutions. What is the product of all these solutions?
7 December
What is the area of the largest triangle that fits inside a regular hexagon with area 952?
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The vertices of this triangle will be every other vertex of the hexagon: and other triangle can be made larger by moving one of its vertices closer to a vertex of the hexagon.
The area of this triangle is 476.
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.
4 December
The last three digits of \(5^5\) are 125.
What are the last three digits of \(5^{2,022,000,000}\)?
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The last three digits of \(5^1\) to \(5^7\) are:
$$
\begin{align*}
5^1&=5&&\rightarrow&&5\\
5^2&=25&&\rightarrow&&25\\
5^3&=125&&\rightarrow&&125\\
5^4&=625&&\rightarrow&&625\\
5^5&=3125&&\rightarrow&&125\\
5^6&=15625&&\rightarrow&&625\\
5^7&=78125&&\rightarrow&&125\\
\end{align*}
$$
This pattern continues: the last three digits of 5 to the power of any odd number (greater than 2) is 125, and the last three digits of 5 to the power of any even number (greater than 3) is 625.
The last three digits of \(5^{2,022,000,000}\) are therefore 625.
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.
2 December
What is the smallest number that is a multiple of 1, 2, 3, 4, 5, 6, 7, and 8?
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8×7×5×3=840. This is also a multiple of 4, 6, 2, and 1. If anything is removed from the product it will no longer be a multiple of all the numbers.
