Puzzles
15 December
The arithmetic mean of a set of \(n\) numbers is computed by adding up all the numbers, then
dividing the result by \(n\).
The geometric mean of a set of \(n\) numbers is computed by multiplying all the numbers together, then
taking the \(n\)th root of the result.
The arithmetic mean of the digits of the number 132 is \(\tfrac13(1+3+2)=2\).
The geometric mean of the digits of the number 139 is \(\sqrt[3]{1\times3\times9}\)=3.
What is the smallest three-digit number whose first digit is 4 and for which the arithmetic and geometric means of its digits are both non-zero integers?
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The only three digit number whose first digits is 4 and for which the arithmetic and geometric means of its digits are both non-zero integers is 444.
Extension
How many three digit numbers are there for which the arithmetic and geometric means of its digits are both non-zero integers?
How many four digit numbers are there for which the arithmetic and geometric means of its digits are both non-zero integers?
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.
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?
23 December
How many numbers are there between 100 and 1000 that contain no 0, 1, 2, 3, or 4?
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These numbers all have 3 digits are there are 5 choices for each digit (5, 6, 7, 8, or 9). The number of numbers is therefore 5×5×5=125.
11 December
There are five 3-digit numbers whose digits are all either 1 or 2 and who do not contain
two 2s in a row: 111, 112, 121, 211, and 212.
How many 14-digit numbers are there whose digits are all either 1 or 2 and who do not contain
two 2s in a row?
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There are two 1-digit numbers whose digits are all either 1 or 2 and who do not contain two 2s in a row (1, and 2).
There are three 2-digit numbers whose digits are all either 1 or 2 and who do not contain two 2s in a row (11, 12, and 21).
There are five 3-digit numbers whose digits are all either 1 or 2 and who do not contain two 2s in a row (111, 112, 121, 211, and 212).
There are eight 4-digit numbers whose digits are all either 1 or 2 and who do not contain two 2s in a row (1111, 1112, 1121, 1211, 1212, 2111, 2112, and 2121).
(It looks like these are the Fibonacci numbers.)
All these numbers either end in a 1, or end in 12. Therefore the \((n+1)\)-digit number can be made by appending a 1 to the end of the \(n\)-digit numbers
or appending a 12 to the end of the \((n-1)\)-digit numbers: so the next term is always the sum of the previous two terms.
Continuing the pattern, we see that there are 987 14-digit numbers are there whose digits are all either 1 or 2 and who do not contain
two 2s in a row.
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?
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.
