(a) The second puzzle for our gambler is a bit similar to the first:
“To paint digits on each side of both dice (one digit per one side) in such a way that any combination from 01 and 31 can be obtained by putting one dice next to the other.”
The digit “6” cannot be used as the digit “9” and vice versa. Is there any solution?
(b) What is the answer to (a) if we allow rotations (i.e. we allow the usage of “6” instead of “9” and vice versa)?
Jane is playing the same game as Kate was playing in Example 3. Can she put together 5 cards with “+” signs and several “digit” cards to make an example on addition with the result equal to 2012
In the following puzzle an example on addition is encrypted with the letters of Latin alphabet: \[{I}+{HE}+{HE}+{HE}+{HE}+{HE}+{HE}+{HE}+{HE}={US}.\] Different letters correspond to different digits, identical letters correspond to identical digits.
(a) Find one solution to the puzzle.
(b) Find all solutions.
(a) After building the garden the successful businesswoman had another idea in mind. She is keen to re-build the terrace in front of her country house. Now the goal is to plant nine sakura trees in such a way that one can count eight rows of trees each consisting of three trees (obviously, a tree can be counted in several rows). How the landscape gardener can satisfy this requirement?
(b) The neighbour of the businesswoman learned about her plans from the talk with the same landscape gardener and decided to outdo her with a similar but more complicated request. He is planning to plant nine sakura trees so that there can be found ten rows of three trees each. Is there a configuration of nine trees satisfying this condition?
Can one cut a square into (a) one 30-gon and five pentagons? (b) one 33-gon and three 10-gons?
A young mathematician had quite an odd dream last night. In his dream he was a knight on a \(4\times4\) board. Moreover, he was moving like a knight moves on the usual chessboard. In the morning he could not remember what was actually happening in his dream, though the young mathematician is pretty sure that either
(a) he has passed exactly once through all the cells of the board except for the one at the bottom leftmost corner, or
(b) he has passed exactly once through all the cells of the board.
For each possibility examine if it could happen or not.
Is it true that if a natural number is divisible by 4 and by 6, then it must be divisible by \(4\times6=24\)?
And what if a natural number is divisible by 5 and by 7? Should it be divisible by 35?
The number \(A\) is not divisible by 3. Is it possible that the number \(2A\) is divisible by 3?
Lisa knows that \(A\) is an even number. But she is not sure if \(3A\) is divisible by 6. What do you think?