Problems

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In a parallelogram \(ABCD\), point \(E\) belongs to the side \(CD\) and point \(F\) belongs to the side \(BC\). Show that the total red area is the same as the total blue area:

In a square, the midpoints of its sides were marked and some segments were drawn. There is another square formed in the centre. Find its area, if the side of the square has length \(10\).

In a parallelogram \(ABCD\), point \(E\) belongs to the side \(AB\), point \(F\) belongs to the side \(CD\) and point \(G\) belongs to the side \(AD\). What is more, the marked red segments \(AE\) and \(CF\) have equal lengths. Prove that the total grey area is equal to the total black area.

Replace the letters with digits in a way that makes the following sum as big as possible: \[SEND +MORE +MONEY.\]

Jane wrote another number on the board. This time it was a two-digit number and again it did not include digit 5. Jane then decided to include it, but the number was written too close to the edge, so she decided to t the 5 in between the two digits. She noticed that the resulting number is 11 times larger than the original. What is the sum of digits of the new number?

a) Find the biggest 6-digit integer number such that each digit, except for the two on the left, is equal to the sum of its two left neighbours.

b) Find the biggest integer number such that each digit, except for the rst two, is equal to the sum of its two left neighbours. (Compared to part (a), we removed the 6-digit number restriction.)

Does there exist an irreducible tiling with \(1\times2\) rectangles of

(a) \(4\times 6\) rectangle;

(b) \(6\times 6\) rectangle?

Irreducibly tile a floor with \(1\times2\) tiles in a room that is

(a) \(5\times8\); (b) \(6\times8\).

Having mastered tiling small rooms, Robinson wondered if he could tile big spaces, and possibly very big spaces. He wondered if he could tile the whole plane. He started to study the tiling, which can be continued infinitely in any direction. Can you help him with it?

Tile the whole plane with the following shapes: