Puzzles

The other three digit numbers that have 3 and 5 as factors and no other prime factors are:

The product is 504.

Valid sequences of length \(n\) can be made by doing one of the following:

Therefore, the number of sequences of length \(n\) that end in B is equal to the total number of sequences of length \(n-1\); and the number of sequences of length \(n\) that end in A is equal to the number of sequences of length \(n-1\) that end in B plus the number of sequences of length \(n-2\) that end in A. Using this we see that:

The total number of valid sequences of length 11 is 638.

29 is odd, and the difference between consecutive square numbers is always odd so it makes sense to check if the relevant two consecutive squares have three digits: these squares are 196 and 225.

The square numbers must be 1, 9, 25, 36, and 784.

The sets of numbers between 1 and \(n\) are either also valid sets with numbers between 1 and \(n-1\); or they can be made by taking a valid set of numbers between 1 and \(n-2\), adding 1 to each number and appending an \(n\) to the set. Therefore the number of sets is the sum of the previous two terms (ie it's the Fibonacci sequence).

Using this, we can work out that the number of sets of numbers between 1 and 14 is 377.

The final digit is 1

Once the leftmost digit of a row and the row below it are decided, there is only one way to fill the rest of the row. Hence, there are two options for the next row (0 0 or 1 1), and four options for the next row (the rows starting 0 and 1 for both 0 0 and 1 1), and in general there are double the number of options for each row as we go upwards.

Therefore for the row with 10 digits there are 2⁹ = 512 options.