Gifted Mathematics

IrMO 2001 P1 Q1: Upper Secondary Mathematics Competition Question

2014-07-13T11:23:00.000+01:00

Find, with proof, all solutions of the equation

2ⁿ = a! + b! + c!

in positive integers a, b, c and n.

(Here, ! means "factorial".)

[IrMO 2001 Paper 1 Question 1]

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Matching Octagons: Middle Secondary Mathematics Competition Question

2014-07-13T11:09:00.000+01:00

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?

Sums of Factorials: Middle Secondary Mathematics Competition Question

2014-07-13T10:46:00.000+01:00

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.

Beads on a Hexagon: Lower Secondary Mathematics Competition Question

2014-07-05T09:05:00.000+01:00

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?

IrMO 2000 P2 Q4: Upper Secondary Mathematics Competition Question

2014-07-05T08:55:00.000+01:00

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]

IrMO 2000 P1 Q5: Upper Secondary Mathematics Competition Question

2014-07-04T07:34:00.000+01:00

Consider all parabolas of the form y = x² + 2px + q (p, q real) which intersect the x- and y-axes in three distinct points. For such a pair p, q let C(p,q) be the circle through the points of intersection of the parabola y = x² +2px+q with the axes. Prove that all the circles C(p,q) have a point in common.

[IrMO 2000, Paper 1, Question 5]

IrMO 2002 P2 Q2: Upper Secondary Mathematics Competition Question

2014-06-27T02:29:00.000+01:00

Suppose n is a product of four distinct primes a, b, c, d such that

(a) a + c = d;

(b) a(a + b + c + d) = c(d - b);

(c) 1 + bc + d = bd.

Determine n.

[IrMO 2002, Paper 2, Question 2]

IrMO 2000 P1 Q3: Upper Secondary Mathematics Competition Question

2014-06-22T07:12:00.000+01:00

Let f(x) = 5x¹³ + 13x⁵ + 9kx. Find the least positive integer k such that 65 divides f(x) for every integer x.

[IrMO 2000, Paper 1, Question 3]

IrMO 1999 P2 Q1: Upper Secondary Mathematics Competition Question

2014-06-20T02:00:00.000+01:00

Solve the system of (simultaneous) equations

y² = (x + 8)(x² + 2);

y² = (8 + 4x)y + 5x² - 16x - 16:

[IrMO 1999, Paper 2, Question 1]

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Domino Products: Lower Secondary Mathematics Competition Question

2014-06-20T01:32:00.000+01:00

Jason is given a set of domino tiles. He is asked to remove all the tiles with a blank square, leaving him with 21 distinct dominoes. He is asked to place two tiles within a 2x2 square grid in such a way that the products of the two numbers along the diagonals are equal.

The diagram shows one solution. In this case, we have 1x4 = 2x2 and we have used the tiles [1,2] and [2,4]. Any solution that is a rearrangement of the same pair of tiles is ignored - we are just interested in which tiles are used.

How many distinct pairs of tiles will solve the above problem?

Primes in a Triangle: Lower Secondary Mathematics Competition Question

2014-06-14T05:55:00.000+01:00

Place nine distinct prime numbers into the grid in such a way that the sums of the four numbers along each side are all equal to each other.

What is the smallest possible such sum?

IrMO 1999 P2 Q2: Upper Secondary Mathematics Competition Question

2014-06-14T01:51:00.000+01:00

A function f : N --> N (where N denotes the set of positive integers) satisfies

(a) f(ab) = f(a)f(b) whenever the greatest common divisor of a and b is 1,

(b) f(p + q) = f(p) + f(q) for all prime numbers p and q.

Prove that f(2) = 2, f(3) = 3 and f(1999) = 1999.

As we're in the year 2014, calculate f(2014).

[adapted from IrMO 1999, Paper 2, Question 2]

Number of Divisors: Middle Secondary Mathematics Question

2014-06-07T08:10:00.000+01:00

Let d(n) be the number of positive divisors of an integer n. For example, d(15) = 4.

Find the smallest positive value of n such that d(n) = d(n+1) = 6.

IrMO 1999, P2 Q4: Upper Secondary Mathematics Competition Question

2014-06-07T07:53:00.000+01:00

Find all positive integers m with the property that the fourth power of the number of (positive) divisors of m equals m.

[IrMO 1999, Paper 2, Question 4]

Three Dice: Upper Primary Mathematics Competition Question

2014-06-01T08:09:00.001+01:00

Trinity rolled three fair six-sided dice, each numbered 1 to 6.

If the sum of her three numbers is 14, what is the highest possible product she could get?

IrMO 1999 P1 Q5: Upper Secondary Mathematics Competition Question

2014-05-31T12:58:00.000+01:00

Three real numbers a, b, c with a < b < c, are said to be in arithmetic progression if c - b = b - a.

Define a sequence u(n), n = 0, 1, 2, 3, ... as follows: u(0) = 0, u(1) = 1 and, for each n > 0, u(n+1) is the smallest positive integer such that u(n+1) > u(n) and {u(0), u(1),... u(n), u(n+1)} contains no three elements that are in arithmetic progression.

Find u(100).

[Irish MO, Paper 1, Question 5, 1999]

Sums of Naturals: Middle Secondary Mathematics Competition Question

2014-05-31T12:30:00.000+01:00

Let N be a natural number with the property that it is the sum of 3 consecutive natural numbers, of 4 consecutive naturals and also of 5 consecutive naturals.

Find the smallest value of N such that the 3 sequences above are disjoint, that is, there is no number that is in more than one sequence.

Pairs of Unit Fractions: Upper Secondary Mathematics Competition Question

2014-05-25T07:35:00.000+01:00

For each positive integer n, let S(n) be the set of ordered pairs (x,y) of positive integers such that

1/n = 1/x + 1/y

and let T(n) be the number of ordered pairs in S(n).

For example, for n=2, S(2)={(3,6), (4,4), (6,3)} and hence T(2)=3.

a) Determine T(n) for all n.

b) Hence, calculate T(2014).

Four Perimeters: Upper Primary Mathematics Competition Question

2014-05-24T05:41:00.001+01:00

A rectangle is divided into four smaller rectangles using two perpendicular lines, as shown in the diagram.

The number inside each small rectangle indicates the length of its perimeter. The diagram is not drawn to scale.

What value should go into the empty rectangle?

Six Eggs in a Box: Lower Secondary Mathematics Competition Question

2014-05-24T05:01:00.000+01:00

Edward likes eggs. Not necessarily to eat them; he likes playing with them too. Today, he has taken eggs from different boxes so that he has 3 white eggs and 3 brown eggs sitting in a 6-egg box. He is thinking about how many patterns he can make with his 6 eggs.

How many different arrangements can Edward make using all 6 eggs in the one box?

As the box has a lid, any similar patterns under rotation count as two distinct arrangements.

Divisibility by 99: Middle Secondary Mathematics Competition Question

2014-05-24T04:28:00.000+01:00

You have nine cards numbered from 1 to 9. If you pick each card randomly and lay them out in order, find the probability that the resulting 9-digit number is divisible by 99.

Express this probability as m/n, where m and n are relatively prime. What is the sum (m + n)?

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A Tangled Peg-Board: Professor Pailyn's Mathematical Quest PMQ46

2013-11-27T03:30:00.001+00:00

Alice was playing around with a peg-board and a jar of elastic bands. She was trying to think up a devilish problem to give Brenda when she finally turned up. Alice had made a real tangled mess and was annoyed with herself for then having to take all the rubber bands off... and then one peg snapped off! She stared at the board, as if willing the peg to jump right back into its rightful place; but it didn't, of course.

However, this now meant that Alice could play with a smaller board. She laid out a perimeter so that she was left with a board of 6 by 6 pins (as shown in the diagram). She had thought of how many elastic bands she'd need if she joined together every pair of pins that were a whole number distance apart, but that had ended up badly. So then she had another idea: join together every pair of pins that are a prime number distance apart. That sounded better! Assuming, of course, that the distance between adjacent pins was a unit length.

So, how many elastic bands will Alice need this time? Do you think she'll have enough of them?!

Note, the diagram shows some examples. Also note that the bands may well overlap each other, but so long as they join two different points that's OK. For example, the horizontal line shown joining 4 pegs contains 3 rubber bands, whereas the vertical line has just 1 band.

Now, where's that Brenda? She's late!

-=0=-

MEMO 2012 Q2: Upper Secondary Mathematics Competition Question

2013-11-21T01:00:00.000+00:00

Let N be a positive integer. A set S is a subset of {1, 2, ..., N} and is called 'allowed' if it does not contain three distinct elements a, b, c such that a divides b and b divides c.

Determine the largest possible number of elements in an allowed set S.

[MEMO 2012 Problem I-2]
[MEMO = Middle European Mathematical Olympiad]

Yin Yang Areas: Lower Secondary Mathematics Competition Question

2013-11-19T01:00:00.000+00:00

The diagram is constructed using one circle of radius 4 units, plus pairs of semicircles of radii 1, 2 and 3 units.

Find the ratio of the red : yellow : green : blue areas.

Integral Network: Professor Pailyn's Mathematical Quest PMQ45

2013-11-15T05:50:00.000+00:00

“That was a bit complicated!” Exclaimed Alice, “But what about my original problem? Can we do that now?”

“I think it’s time to recharge those brain cells... with some tea and fruit cake.” Professor Pailyn walked straight past Alice, carrying the tea-tray into the garden.

Alice enjoyed elevenses; it was almost better than breakfast. Alice had never had a proper stroll around the Professor’s garden. As she walked through it with her eyes, she noticed some peculiar structures: a cube within a cube standing on one corner, and what appeared to be a leafy halo suspended in mid-air betwixt two egg-shapes. She wanted to go and find the hidden wires but was too busy eating cake.

“So... let’s have a little think about your problem. You have four points on a plane, all connected to each other by straight lines. No three points lie on the same straight line so that we have six distinct line segments. You want all six lines to be different whole numbers. Is that right?”

“Yes, that’s right.” Said Alice. “And what are the smallest possible distances?”

“Do you mean the smallest sum of the distances or the minimum value of the largest distance?” asked Professor Pailyn. Alice looked unsure. “They may be the same solution, so let’s try to find the minimum largest distance.” Alice felt relieved. “Now, if we were allowed to have three points in a straight line, then that solution is fairly easy to find. And if we were allowed to have some lengths equal to each other, that too is doable, even if a bit harder. Actually, placing the points at the corners of a Pythagorean rectangle is the easiest of all. But this is trickier.”

Alice was getting used to Professor Pailyn’s mischievous smile. He had never set her anything that had no solutions, but she was always vigilant against being led down the garden path. Mind you, in this case, this was her problem.

-=o0o=-