Problems

Age
Difficulty
Found: 32

Illustrate with a picture

(a) \((a-b)^2 = a^2 - 2ab + b^2\),

(b) \(a^2 - b^2 = (a-b)(a+b)\),

(c) \((a+b+c)^2 = a^2 + b^2 + c^2 + 2ab + 2ac + 2bc\).

Suppose \(a>b\). Explain using the number line why

(a) \(a-c>b-c\), (b) \(2a>2b\).

Suppose \(a>b\) and \(c>d\). Prove that \(a+c>b+d\).

Using mathematical induction prove that \(2^n \geq n + 1\) for all natural numbers.

Circles and lines are drawn on the plane. They divide the plane into non-intersecting regions, see the picture below.

Show that it is possible to colour the regions with two colours in such a way that no two regions sharing some length of border are the same colour.

Consider a number consisting of \(3^n\) digits, all ones, such as 111 or 111111111 for example. Show that such a number with \(3^n\) digits is divisible by \(3^n\).

Numbers \(1,2,\dots,n\) are written on a whiteboard. In one go Louise is allowed to wipe out any two numbers \(a\) and \(b\), and write their sum \(a+b\) instead. Louise enjoys erasing the numbers, and continues the procedure until only one number is left on the whiteboard. What number is it? What if instead of \(a+b\) she writes \(a+b-1\)?

Prove that

(a) \[1^2 + 2^2 + 3^2 + \dots + n^2 = \frac{1}{6} n (n+1)(2n+1)\]

(b) \[1^2 + 3^2 + 5^2 + \dots + (2n-1)^2 = \frac{1}{3} n (2n-1)(2n+1)\].

With a non-zero number, the following operations are allowed: \(x \rightarrow \frac{1+x}{x}\), \(x \rightarrow \frac{1-x}{x}\). Is it true that from every non-zero rational number one can obtain each rational number with the help of a finite number of such operations?

On a particular day it turned out that every person living in a particular city made no more than one phone call. Prove that it is possible to divide the population of this city into no more than three groups, so that within each group no person spoke to any other by telephone.