Puzzles

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.

The product is 120.

In order for the number of odd factors of the number to be odd, it must be an odd square number or a power of two times an odd square number. 128 is 1 (an odd square) times 128 (a power of two) and no smaller three digit number is of this form.

A valid sequence of length \(n\) can be made by either adding a single B to the right of a valid sequence of length \(n-1\) or two As to a valid sequence of length \(n-2\). Therefore, the number of sequences of length \(n\) can be calculated by adding the previous two terms (ie it's the Fibonacci sequence).

Using this, we can work out that the number of sequence of length eleven is 144.

3 down is a multiple of 101, so its middle digit must be 0:

The clue 4 across now tells us that the units digit of today's number (ie 1 across) is 5 or 0. As 3 down cannot begin with 0, it must be 5.

3 down is a multiple of 101:

The sum of the digits of today's number is at most 9+9+5 = 23, so 1 down starts with a 1 or 2. This means that the sum of the digits of today's number is at most 2 + 9 + 5 = 16, so 1 down must start with a 1:

Two times today's number (ie 4 across) must start with a 2 or 3. If it started with 2, then the sum of the digits of today's number (1 down) would be 12, so today's number would be 165. The product of the digits of today's number (5 across) would be 30. This is not possible as 5 across ends in 5. The first digit of 4 across must therefore be a 3:

The sum of the digits of today's number is 13, and we can fill in the rest using the clues for 4 and 5 across:

Therefore, today's number is 175.

The sum of the units digits of the three numbers must be 10 (as the units digit of the result must be 0 and the number are not big enough to allow the sum to be 20). Similarly, the sum of the two tens digits must be 9, and the sum of the two hundreds digits is 9 (so that 10s are made when the 1 is carried).

The> hundreds digits could be 2 and 7, 3 and 6, or 4 and 5. As we're looking to make the larger number as small as possible, we should pick 4 and 5.

The tens digits could be 2 and 7 or 3 and 6. We pick 2 and 7 and as 2 is the smallest possible number here.

Finally, the units digits are 1, 3 and 6. We use 1 for our number as it's smallest, leading to 521.

The sum of \(n\) consecutive numbers starting at \(m\) is \(m + (m + 1) + ... + (m + n-1) = nm + 0 + 1 + ... + n - 1\). The sum of the next \(n\) consecutive number is \((m+n) + (m+n+1) + ... + (m + 2n - 1 ) = n(m + n) + 0 + 1 + ... + n - 1\). The difference between these two is \(n(m + n) - nm = n^2\).

Therefore, whatever value of \(m\) is used, \(n\) will always be 451.

If the numbers \(p_1\), \(p_2\), ..., \(p_n\) are odd prime numbers and \(i_1\), \(i_2\), ..., \(i_n\) are positive integers, then the number \(p_1^{i_1}p_2^{i_2}...p_n^{i_n}\) has \((i_1+1)(i_2+1)...(i_n+1)\) odd factors.

If \((i_1+1)(i_2+1)...(i_n+1)=16\) the the possible values for the \(i\)s are:

These options lead to: