Puzzles

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

By rotating the triangles, they can be made to exactly cover the decagon.

The total area of all the triangles is therefore 240.

The first division tells that 12345 is 205 more than a mutliple of today's number, and so today's numwber is a factor of 12140.

The second division tells that 6789 is 112 more than a mutliple of today's number, and so today's number is a factor of 6677.

The only numbers that are factors of both 12240 and 6677 are 1 and 607.

The largest possible isosceles triangles with a perimeter of 50 units is the equilateral triangle with sides of length 50/3. The area of this triangle is 120.28 square-units.

By continuously adjusting the width of the base of the isosceles triangle, triangles with every area between 0 and 120.28 can be created. The animation below shows this.

Triangles with areas of each integer from 1 to 120 can therefore be created. Each area can be made twice: once with a tall and thin triangle, and once with a short and wide triangle. This means that there are 240 different triangles with a perimeter of 50 units and an integer area.

The product of the numbers in the red boxes is 160.

The numbers 10, 20, ..., 1000 end with a 0. There are 100 of these numbers.

The numbers 100, 101, ..., 109, 200, ..., 209, 300, ..., 309, ..., 908, 909, 1000 contain a 0 in their tens column. There are \(10\times9+1=91\) of these numbers.

The number 1000 has a 0 in its hundreds column.

In total, there are 100+91+1=192 zeros.