MO Schulrunde Klasse 9 Mock Paper 4 with Solutions

MO Schulrunde Klasse 9 Mock Paper 4 · IMOolympiad.com · Original practice

6 written-solution problems · Take-home practice: choose any 4 of 6 problems

For school year 9. The official shared school-round sheet for years 9–10 offers six problems; the local organiser decides which problems and working arrangements apply. Our practice uses the choose-four option.

Original independent practice in English. Not an official paper or predicted selection test. Keep hints and solutions closed during your attempt.

How to review your proof

Check that you have used every condition, justified the main idea, covered all cases and stated the conclusion. A different complete proof can also be correct. This is a self-review checklist; written proofs are not automatically marked.

Question 1

Prove that an integer n is odd if and only if $n^2$ leaves remainder 1 when divided by 8.

Hint 1

For odd n write n=2k+1.

Hint 2

Of two consecutive integers k,k+1, one is even.

Worked solution 1

If n=2k+1, then $n^2-1=4k(k+1)$ . Since k(k+1) is even, this is divisible by 8, so the remainder is 1. Conversely, if n were even, write n=2k. Its square is 4k², whose remainder modulo 8 is 0 or 4 according as k is even or odd. It cannot be 1. Both directions are proved.

Conclusion: Exactly the odd integers have square remainder 1 modulo 8.

Review the idea: Parity · Remainders

Question 2

Determine all integers x for which $x^2+3x+2$ is a perfect square. Zero is allowed as a perfect square.

Hint 1

Factor the expression as $(x+1)(x+2)$ .

Hint 2

The product of two consecutive positive integers lies strictly between their squares.

Worked solution 2

Put t=x+1, so the expression is t(t+1). For t at least 1, $t^2\lt t(t+1)\lt (t+1)^2$ , so it is not a square. For t at most -2, put u=-t-1, which is at least 1; then t(t+1)=u(u+1), also not a square. The remaining cases t=0 and t=-1 both give zero. They correspond to x=-1 and x=-2, and both work.

Conclusion: Exactly x=−2 and x=−1.

Review the idea: Inequality rules and signs · Factorisation integer solutions

Question 3

A standard 8-by-8 square board has its top-left and bottom-right corner cells removed. Prove that the remaining board cannot be covered by 1-by-2 dominoes without gaps or overlaps. Dominoes must follow the grid.

Hint 1

Colour the board in the usual alternating two colours.

Hint 2

The removed corners have the same colour.

Worked solution 3

Originally the board has 32 cells of each colour. The two opposite corners have the same colour, so after removal there are 30 cells of one colour and 32 of the other. Every grid-aligned domino covers two neighbouring cells, hence one of each colour. Any union of disjoint dominoes has equal colour counts, which the remaining board does not. Therefore such a covering is impossible.

Conclusion: Impossible by unequal colour counts.

Review the idea: Parity · Proof methods

Question 4

ABCD is a square with vertices in boundary order. Equilateral triangle ABE is constructed on side AB outside the square. Find $\angle DEC$ , proving your answer.

Hint 1

Compare the equal sides AD and AE.

Hint 2

Angles DAE and CBE are 150 degrees.

Worked solution 4

Since E is outside across AB, $\angle DAE=90^\circ+60^\circ=150^\circ$ . The lengths AD and AE both equal the square side, so triangle ADE is isosceles and $\angle AED=(180^\circ-150^\circ)/2=15^\circ$ . Similarly $\angle CEB=15^\circ$ . The rays ED and EC lie between EA and EB in that order, as follows from the convex square and the exterior position of E. Thus $\angle DEC=\angle AEB-\angle AED-\angle CEB=60^\circ-15^\circ-15^\circ=30^\circ$ .

Conclusion: Angle DEC is 30°.

Review the idea: Angles · Triangle congruence

Question 5

Prove that $n^2+n+41$ is composite for infinitely many positive integers n.

Hint 1

Choose n to be a positive multiple of 41.

Hint 2

One factor is then 41, and the other is greater than 1.

Worked solution 5

For any positive integer k, put n=41k. Then $n^2+n+41=41(41k^2+k+1)$ . Both integer factors exceed 1: the first is 41, and the second is at least 43. Thus the value is composite. Distinct positive k give distinct positive n, so this supplies infinitely many required integers.

Conclusion: Every positive multiple n=41k gives a composite value.

Review the idea: Divisibility

Question 6

Triangle ABC is right-angled at A. Let M be the midpoint of BC. If $\angle BAM=20^\circ$ , determine angles B and C.

Hint 1

The midpoint of a right triangle’s hypotenuse is equidistant from its three vertices.

Hint 2

Use the isosceles triangle ABM.

Worked solution 6

The circle with diameter BC passes through A, so its centre M satisfies MA=MB. Triangle ABM is therefore isosceles, and angle ABM equals angle BAM, namely 20 degrees. Since B,M,C are collinear in that order, angle ABC is also 20 degrees. The remaining angle is $180^\circ-90^\circ-20^\circ=70^\circ$ .

Conclusion: B=20°, C=70°.

Review the idea: Circles and power of a point · Angles

After this paper

Choose one gap in your proof to repair, study the linked idea, and write a complete solution again before the next mock.

Choose another paper · Check your German selection route

Format reference: official organiser information. Questions and explanations are independent practice material.