Puzzles

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:

The line passes through one square in each column plus an extra square in a column whenever is passes between rows. It must pass between rows 250 times (to get from row 1 to 251) so in total it passes through 250+272 = 522 squares.

In a 200 by 300 grid of squares, how many squares will a line drawn from the bottom left corner to the top right corner pass through?

For a multiplication table from 1 to \(n\), the sum of all the numbers in the top row is \(1+...+n\), the sum of all the numbers in the second row is \(2(1+...+n)\), the sum of all the numbers in the third row is \(3(1+...+n)\), ..., and the sum of all the numbers in the \(n\)th row is \(n(1+...+n)\). The sum of all the numbers is therefore \((1+...+n)(1+...+n)=n^2(n+1)^2/4\) and the sum of the numbers in the bottom row is \(n^2(n+1)/2\).

If the sum of all the numbers is 2025, then \(n=9\) and so the sum of the numbers in the bottom row if 405.

In each match, one person is knocked out, so for a tournament with \(n\) players, you'll always need \(n-1\) matches (as everyone except the winner must be knocked out). Therefore there will be 354 matches however Carol arranges the byes.

The product is 810.

We know the answer has three digits: we are looking for a number ABC that is equal to 4×A×B×C. Dividing by A, we see that 4×B×C > 100 and so B×C > 25. This means that the only possible options for BC are 65, 66, 74, 75, 76, 77, 84, 85, 86, 87, 88, 93, 94, 95, 96, 97, 98 and 99.

Trying these options with different values for A, we find that 384 works.