Puzzles

The product of the numbers in the red boxes is 144.

The rightmost numbers in each row are the square numbers. The smallest square number larger than 300 is 324.

There are \((729-n)/2\) even numbers between \(n\) and 729, and so we want to solve \((729-n)/2=n\). The solution of this is 243.

For which odd numbers \(N\) does there exist an odd number \(n\) such that there are \(n\) even numbers between \(n\) and \(N\)?

The largest deteminant will be made by making \(a\) and \(d\) as large as possible (ie 12) and \(b\) and \(c) as small as possible (ie 1). This gives a determinant of 143.

The determinant of the 3 by 3 matrix \(\begin{pmatrix}a&b&c\\d&e&f\\g&h&i\end{pmatrix}\) is \(a(ei-fh)-b(di-fg)+c(dh-eg)\).

If a 3 by 3 matrix's entries are all in the set \(\{1, 2, 3\}\), the largest possible deteminant of this matrix is 28.

What is the largest possible determinant of a 3 by 3 matrix whose entries are all in the set \(\{1, 2, 3, ..., 12\}\)?

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.

The largest number you can make with the digits in the red boxes is 532.

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.

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?

The product of the numbers in the red boxes is 315.