Problems

Age
Difficulty
Found: 2449

Find all the solutions of the puzzle and prove there are no others. Different letters denote different digits, while the same letters correspond to the same digits. \[M+MEEE=BOOO.\]

Philip and Denis cut a watermelon into four parts. When they finished eating watermelon (they ate the whole thing), they discovered that there were five watermelon rinds left. How is it possible, if no rind was cut after the initial cutting?

Place coins on a \(6\times 6\) chequered board (one coin on one square), so that all the horizontal lines contain different number of coins, and all vertical lines contain the same number of coins.

Cut a square into a heptagon (7 sides) and an octagon (8 sides) in such a way, that for every side of an octagon there exists an equal side belonging to the heptagon.

Is it possible to arrange some group of distinct numbers in a circle so that each number equals the sum of its neighbours?

A big square was cut into smaller squares. Sebastian used all the pieces and constructed two squares with different side lengths by glueing the pieces together. Show an example of how he could do that.

Sarah is writing down natural numbers starting from 2. She notices that each time she writes the next number the sum of all written numbers is less than their product. She believes she can find such 57 natural numbers (not necessarily different from each other) that their sum will be greater than their product. Do you think it is possible?

(a) Can you find a set of distinct numbers which can be arranged in a circle in such a way that each number equals the product of its neighbours?

(b) Is it true that each solution of Example 1 is determined by the values of two neighbouring numbers?

It was Sebastian’s younger brother who cut the big square in Example 2. Now you need to help him to cut one of the squares (which Sebastian obtained after glueing the pieces) into smaller congruent triangles. But please make sure the elder brother can do the same thing as before: to divide the resulting congruent triangles into two groups and to glue the pieces of each group together to make two squares with different side lengths.

It’s not that difficult to find a set of \(57\) integers which has a product strictly larger or strictly smaller than their sum. Is it possible to find \(57\) integers (not necessarily distinct) with their sum being equal to their product?