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?
George divided number \(a\) by number \(b\) with the remainder \(d\) and the quotient \(c\). How will the remainder and the quotient change if the dividend and the divisor are increased by a factor of 3?
Let us introduce the notation – we denote the product of all natural numbers from 1 to \(n\) by \(n!\). For example, \(5!=1\times2\times3\times4\times5=120\).
a) Prove that the product of any three consecutive natural numbers is divisible by \(3!=6\).
b) What about the product of any four consecutive natural numbers? Is it always divisible by 4!=24?