2

^{n}= a! + b! + c!

in positive integers a, b, c and n.

(Here, ! means "factorial".)

[IrMO 2001 Paper 1 Question 1]

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Find, with proof, all solutions of the equation

2^{n} = a! + b! + c!

in positive integers a, b, c and n.

(Here, ! means "factorial".)

[IrMO 2001 Paper 1 Question 1]

2

in positive integers a, b, c and n.

(Here, ! means "factorial".)

[IrMO 2001 Paper 1 Question 1]

The image shows two regular octagons, each has 4 red and 4 blue beads, one placed on each vertex.

We say that there is a 'match' if a vertex has the same colour in both octagons. In the diagram, we can see that the upper-right vertices are both blue and the lower-right vertices are both red - all the other vertex pairs have different colours. In this case, we have a matching value of 2.

However, if we rotate the inner octagon by 45 degrees clockwise then all the vertices match, giving us a maximum matching score of 8 for these two arrangements.

Now, if we randomly allocate the beads to the two octagons (each with 4 of each colour) and we are allowed to rotate one of the octagons, what is the smallest*guaranteed *maximum matching score?

We say that there is a 'match' if a vertex has the same colour in both octagons. In the diagram, we can see that the upper-right vertices are both blue and the lower-right vertices are both red - all the other vertex pairs have different colours. In this case, we have a matching value of 2.

However, if we rotate the inner octagon by 45 degrees clockwise then all the vertices match, giving us a maximum matching score of 8 for these two arrangements.

Now, if we randomly allocate the beads to the two octagons (each with 4 of each colour) and we are allowed to rotate one of the octagons, what is the smallest

Let N = a! + b!

Find all solutions (a,b) such that N is divisible by 11 and both a and b are positive integers less than 11, with a ≤ b.

How many solutions are there?

You may leave your answers in the form n!, where n! = n.(n-1).(n-2)... 3.2.1.

Six beads are arranged at the corners of a regular hexagon; 3 are orange and 3 green. All arrangements that are rotational symmetries of each other count as one unique arrangement.

Using all six beads, how many unique arrangements are there?

Prove that in each set of ten consecutive integers there is one which is coprime with each of the other integers.

For example, taking 114, 115, 116, 117, 118, 119, 120, 121, 122, 123 the numbers 119 and 121 are each coprime with all the others. [Two integers a, b are coprime if their greatest common divisor is one.]

[IrMO 2000 Paper 2 Question 4]

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