Puzzles

The mean of the numbers in the \(n\)th row is \(n^2\), so the mean of the number in the 19th row is 361.

The route must contain three upward movements. As the same line cannot be travelled along twice, there is only one possibility for the rest of the route. There are 8 choices for the position of each upward movement, so there are 8×8×8=512 possible routes.

The two triangles are similar, and the sides of the blue triangle are twice the length of the red triangle. Therefore the area of the blue triangle is 4 times the area of the red triangle. 226×4=904.

The path from A to B will consist of 10 moves. 3 of these moves must be moves up. Therefore there are "10 choose 3" possible routes. 10 choose 3 is 120.

The rightmost number is the \(n\)th row is \(n^2\). 750 is between \(27^2\) and \(28^2\) so is in the 28th row.

The numbers in the \(n\)th row are \(2n\) less than the numbers below them, and so the number above 750 is 2×27 less than 750. 750–2×27=696.

The largest number you can make using the digits in the red boxes is 654.

If Eve had written down an even number of numbers, then the mean of her numbers would be halfway between the two middle integers, so would be an integer and a half.

If Eve had written down two numbers, then the mean of those two numbers would be 422. This is not an integer and a half.

If Eve had written down three numbers, then the mean of those three numbers would be 281.3333. This is not an integer.

If Eve had written down four numbers, then the mean of those two numbers would be 211. This is not an integer and a half.

If Eve had written down five numbers, then the mean of those three numbers would be 168.8. This is not an integer.

If Eve had written down six numbers, then the mean of those two numbers would be 140.6666 This is an not integer and a half.

If Eve had written down seven numbers, then the mean of those two numbers would be 120.5714. This is an not integer.

If Eve had written down eight numbers, then the mean of those two numbers would be 105.5. This is an integer and a half, so this is possible. The eight consecutive integers with a mean of 105.5 are 105 and 106, 104 and 107, 103 and 108, and 102 and 109.

To be certain that this is the only answer, we should rule out Eve having written more than eight numbers.

The sum of the numbers from 1 to 41 is 861. This is greater than 844, so Eve must have written fewer than 41 numbers.

If Eve had written down an odd number of numbers, then the mean of her numbers would be an integer. This means that the number of numbers would be a factor of the sum of all the numbers. 844's prime factorisation is 2×2×211, and so its only odd factor is 211. This is greater than 41, so Eve didn't write down this many numbers.

Eve therefore must have written down an even number of numbers.

If Eve wrote down an even number of numbers, then the mean of her numbers would be an integer plus a half. This implies that half the number of numbers would be a factor of the sum, and the number of numbers cannot be a factor of the sum. This is only possible if there are 8 numbers.

836=2×2×11×19, and so for the product to be correct the integers could be:

The only one of these whose sum is 51 are 2, 11, and 38. Therefore the product of the largest and second-largest integers is 418