MO Landesrunde Klasse 9 Mock Paper 4: Questions, Hints and Solutions

Write $n=2^a p_1^{e_1}\cdots p_r^{e_r}$, where the p’s are distinct odd primes and a is a nonnegative integer; the product may be empty. Every divisor is obtained by independently choosing each prime exponent, so $$\sigma(n)=(1+2+\cdots+2^a)\prod_{i=1}^r(1+p_i+\cdots+p_i^{e_i}).$$ The first factor is odd. Each term in the factor for an odd prime is odd, so that factor is odd exactly when its number of terms, $e_i+1$, is odd, or $e_i$ is even. Therefore the divisor sum is odd precisely when all odd-prime exponents are even. If a is also even, n is a square; if a is odd, n is twice a square. Conversely either of those two forms has all odd-prime exponents even and therefore an odd divisor sum. This includes n=1.

Conclusion: Exactly squares and twice squares.

Review the idea: Unique prime factorisation · Divisibility

Set $x=(b+c-a)/2$, $y=(c+a-b)/2$, and $z=(a+b-c)/2$. Triangle inequalities make x,y,z positive, with a=y+z, b=z+x and c=x+y. The left side is 8xyz, while the right side is $(x+y)(y+z)(z+x)$. The inequality $(\sqrt x-\sqrt y)^2\ge0$ gives $x+y\ge2\sqrt{xy}$, and similarly for the other pairs. Thus each pair sum is at least twice the square root of its product. Multiplying gives $(x+y)(y+z)(z+x)\ge8xyz$, proving the result. Equality requires x=y=z, which is equivalent to a=b=c. Conversely an equilateral triple gives equality.

Conclusion: Equality exactly for an equilateral triangle.

Review the idea: Triangle inequalities · Arithmetic geometric and harmonic means

For a circle with centre (u,v) and radius r, define the power of X=(x,y) as $(x-u)^2+(y-v)^2-r^2$. For two distinct circles that intersect, equality of their powers is a nonzero linear equation: the x²+y² terms cancel, and their centres cannot coincide. The equation holds at both intersection points, so its set of solutions is exactly their common-chord line. At P, lying on two common-chord lines, the powers for the three circles are equal by transitivity. Therefore P satisfies the equality for the remaining pair and lies on its common-chord line. If two of the common-chord lines coincide, the same transitivity holds for every point on that line and the third line coincides too.

Conclusion: The three lines are concurrent, or all coincide when two do.

Review the idea: Circles and power of a point · Equations and quadratics

In each column, at least two of the three cells share a colour. Select one such pair and record its two row numbers and colour. There are $\binom32\cdot2=6$ possible records. If m≥7, two columns have the same record, yielding a forbidden monochromatic rectangle. Thus m≤6. For attainability, take exactly the six different columns containing both colours: RRB,RBR,BRR,RBB,BRB,BBR. Each has exactly one same-colour row-pair, and those six records are all different. A monochromatic rectangle would repeat a record, so none occurs. Hence the maximum is 6.

Conclusion: Maximum m=6.

Review the idea: Pigeonhole principle · Counting

Because the last two fractions are positive, x cannot be 1. Also the ordering gives $1\le3/x$, so x is 2 or 3. If x=3, both remaining fractions are at most 1/3, so equality in their sum forces y=z=3. If x=2, then $1/y+1/z=1/2$, or $yz=2y+2z$. Thus $(y-2)(z-2)=4$. Neither y nor z can be 2, since the other reciprocal would then have to be zero. The positive factors y−2 and z−2 are ordered, so they are (1,4) or (2,2). We obtain (2,3,6) and (2,4,4). Direct substitution verifies these and (3,3,3), completing both directions.

Conclusion: Exactly $(2,3,6),(2,4,4),(3,3,3)$.

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

All terms are integers at least 2. We first prove the product formula $a_{n+1}-1=a_1a_2\cdots a_n$. It holds at n=1, because a₂−1=2. If it holds at n−1, then $a_{n+1}-1=a_n(a_n-1)=a_n a_1\cdots a_{n-1}$, proving it by induction. Thus any later term is 1 modulo any earlier term, so the terms are pairwise coprime. Also $$\frac1{a_k-1}-\frac1{a_{k+1}-1}=\frac1{a_k-1}-\frac1{a_k(a_k-1)}=\frac1{a_k}.$$ Summing telescopes to $1/(a_1-1)-1/(a_{n+1}-1)=1-1/(a_{n+1}-1)$, as claimed.

Conclusion: The terms are pairwise coprime, and the stated reciprocal sum identity holds.

Review the idea: Recursive sequences · Greatest common divisor · Sequences and sums