Easiest IMO problems that will make you feel like a Genius


IMO (International Mathematics Olympiad) is known for its tough problems that make people sweat but occasionally, easy problems have been provided across the years. These are problems where you do not need even a paper to solve them. Your imagination is enough for such simple problems. We will go through such problems so that we feel like a GENIUS for good.

Some of the easiest problems that came in IMO (International Mathematics Olympiad) are as follows:

  • IMO 1959 Problem 1: Solved using Euclid's algorithm
  • IMO 1964 Problem 1: Solved using simple modulus
  • IMO 1984 Problem 1: Solved using AM GM inequality
  • IMO 1986 Problem 1: Solved using simple modulus
  • IMO 2012 Problem 2: Solved using AM GM inequality

We will go through these problems one by one and it is a guarantee that this will make you feel that you will ace IMOs and emerge as the next champion.

IMO 1959 Problem 1 (Romania)

For every integer N, prove:
(21N+4) / (14N+3) cannot be reduced further.

Simple solution:

Solution 1:

If the fraction is irreducible it means the denominator and numerator have no common divisors.

Assuming k divides 21n+4 and 14n+3, then k also divides 2(21n+4) = 42n + 8 and 3(14n+3)=42n+9 as well as the difference (42n+9)-(42n+8)=1 - we conclude that the only common divisor is 1.

Solution 2:

Euclid's Algorithm to find the Greatest Common divisor between two numbers say A and B (where A >= B) is as follows:

GCD(A, B) = GCD(B, A-B)
till the point B > 1

By Euclid's Algorithm:

  gcd(21n+4, 14n+3) 
= gcd(14n+3, 7n+1)
= gcd(7n+2, 7n+1)
= gcd(7n+1, 1)
= 1

Hence, proved.

IMO 1964 Problem 1 (Moscow)

The problem statement is as follows:

  • Find all positive integers N for which 2^N - 1 is divisible by 7.

  • Prove that there is no positive integer N for which 2^N + 1 is divisible by 7.

The solution is simple.

When 2^N is divided by 7, there can be only 3 different remainders.

$$ 2^N mod 7 = 2 or 4 or 1 $$

With this, we get the following:

$$ (2^N - 1) mod 7 = 1 or 3 or 0 $$

As we are getting 0 as a remainder for 2^N-1, this means that for certain values of N, 2^N-1 is perfectly divisible by 7.

Such values are where N is a multiple of 3. For example:

2^3 = 8 ; 2^6 = 64 ; ...

Let us now move to the second part of the problem.

$$ (2^N + 1) mod 7 = 3 or 5 or 2 $$

As we do not get 0 as a remainder, we know that that (2^N+1) is never perfectly divisible by 7.

Enjoy. You solved this problem like Butter.

IMO 1984 Problem 1 (Prague, Czechoslovakia)

Prove that 0 ≤ yz + zx + xy − 2xyz ≤ 7/27, where x, y and z are non-negative real numbers for which x + y + z = 1.

Solution:

(1 - 2x)(1 - 2y)(1 - 2z) 
= 1 - 2(x + y + z) + 4(yz + zx + xy) - 8xyz 
= 4(yz + zx + xy) - 8xyz - 1. 

Hence:

yz + zx + xy - 2xyz 
= 1/4 (1 - 2x)(1 - 2y)(1 - 2z) + 1/4. 

By the arithmetic/geometric mean theorem 

(1 - 2x)(1 - 2y)(1 - 2z) ≤ ((1 - 2x + 1 - 2y + 1 - 2z)/3)^3

 = 1/27. 
 
So yz + zx + xy - 2xyz ≤ 1/4 28/27 = 7/27. 

IMO 1986 Problem 1 (Warsaw, Poland)

Let d be any positive integer not equal to 2, 5, or 13. Show that one can find distinct a, b in the set {2, 5, 13, d} such that ab − 1 is not a perfect square.

Solution:

Consider residues mod 16.

A perfect square must be 0, 1, 4 or 9 (mod 16). 

d must be 1, 5, 9, or 13 for 2d - 1 to have one of these values. 

However:

  • if d is 1 or 13, then 13d - 1 is not one of these values.
  • If d is 5 or 9, then 5d - 1 is not one of these values.

So we cannot have all three of 2d - 1, 5d - 1, 13d - 1 as perfect squares.

IMO 2012 Problem 2 (Mar del Plata, Argentina)

The problem is as follows:

Let a_2, a_3, ..., a_n be real positive numbers that statisfies the following relation: (a_2) * (a_3) * ... * (a_n) = 1. We need to prove that:

$$ (a_2 + 1)^2 * (a_3 + 1)^3 ... * (a_n + 1)^n >= n^n $$

This problem can be solved easily using ideas of Arithmetic Mean and Geometric Mean.

The AM GM inequality states that:

$$ (x_1 + x_2 + ... + x_n) / n >= ( x_1 * x_2 * ... * x_n )^(1/n) $$

The problem is easily solved with this result.

Let us apply the inequality for (a_k + 1)^k.

$$ (a_k + 1)^k = (a_k + 1/(k-1) + 1/(k-1) + ... + 1/(k-1))^k $$

$$ (a_k + 1)^k >= k^k * (a_k * 1/(k-1) * 1/(k-1) * ... * 1/(k-1))^(k/k) $$

$$ (a_k + 1)^k >= k^k * (a_k * 1/(k-1) * 1/(k-1) * ... * 1/(k-1)) $$

$$ (a_k + 1)^k >= k^k * (a_k) / (k-1)^(k-1)

Hence, we get the central result. Now, if we use this to evaluate the left hand side of the original equation, we get the result we are looking for.

$$ LHS = (a_2 + 1)^2 * (a_3 + 1)^3 ... * (a_n + 1)^n $$

$$ LHS >= ((a_2) * (2^2) / (1^1)) * ... * ((a_n) * (n^n) / (n-1)^(n-1)) $$

$$ LHS >= n^n * (a_2 * a_3 * ... * a_n) $$

$$ LHS >= n^n $$

as it is given in the problem that a_2 * a_3 * ... * a_n = 1.

Congratulations, you solved yet another IMO problem easily. You are a genius.

Celebrate your victory with this:

elon_1-1