MO Bundesrunde Klasse 9 Mock Paper 5: Questions, Hints and Solutions

Write $N=2^{s-1}m$, with m odd and s≥2. If $\sigma(m)$ denotes the sum of m’s positive divisors, every divisor of N is uniquely a power of 2 times a divisor of m. Thus $\sigma(N)=(2^s-1)\sigma(m)$. Perfection is equivalent to $$(2^s-1)\sigma(m)=2^s m.$$ Since consecutive integers $2^s-1$ and $2^s$ are coprime, $2^s-1$ divides m. Write $m=q(2^s-1)$; cancellation gives $\sigma(m)=2^s q=m+q$. If q>1, then 1,q,m are distinct divisors of m, because 2ˢ−1≥3. Their sum already exceeds m+q, a contradiction. Hence q=1 and $\sigma(m)=m+1$, which means m is prime: a composite m>1 has an additional proper divisor. This proves the required form. Conversely, if $p=2^s-1$ is prime, its divisor sum is p+1=2ˢ, so $\sigma(2^{s-1}p)=(2^s-1)2^s=2N$. The proposed number is indeed perfect.

Conclusion: Exactly the even numbers of the stated form with 2ˢ−1 prime.

Review the idea: Counting and summing divisors · Unique prime factorisation

Put f(x)=$x^{3}$−3x. For x<y, $f(y)-f(x)=(y-x)(x^2+xy+y^2-3)$. If both x,y lie in [1,∞), the second factor is positive for distinct x,y; the same is true on (−∞,−1]. If both lie in [−1,1], that factor is negative for distinct x,y: each of $x^{2}$,$y^{2}$,xy is at most 1, and equality of their sum to 3 would force x=y=1 or x=y=−1. Thus f is strictly increasing on the two outer intervals and strictly decreasing on the middle one. Polynomials are continuous; with f(−1)=2, f(1)=−2 and unbounded outer values, their respective ranges are (−∞,2], [−2,2], and [−2,∞). Consequently there are three distinct roots for −2<t<2, one for |t|>2, and two when t=±2. At t=2, $x^3-3x-2=(x-2)(x+1)^2$, giving −1 and 2. At t=−2, $x^3-3x+2=(x+2)(x-1)^2$, giving −2 and 1.

Conclusion: Three roots for |t|<2; two for |t|=2; one for |t|>2.

Review the idea: Polynomial functions · Solving polynomial equations

Place M at (0,0) and PQ on the horizontal axis. If X and Y have horizontal coordinates ξ and η, the goal is to prove ξ+η=0. Write P=(−r,0), Q=(r,0), with r>0. The centre lies on their perpendicular bisector, so it is (0,h) for some real h, and the circle equation is $x^2+y^2-2hy=r^2$. Let $u=(u_1,u_2)$ and $v=(v_1,v_2)$ be unit direction vectors along AB and CD, meaning $u_1^2+u_2^2=v_1^2+v_2^2=1$. Write A=a u, B=b u, C=c v, D=d v, using signed real numbers a,b,c,d. For example $a u=(au_1,au_2)$, and a negative a means the direction opposite u. None is zero because M is inside the circle. Substituting a point t u into the circle gives $t^2-2hu_2t-r^2=0$, whose roots are a,b. Hence $$\begin{gathered}a+b=2hu_2,\\ ab=-r^2,\\ \frac1a+\frac1b=-\frac{2hu_2}{r^2}.\end{gathered}$$ Similarly $1/c+1/d=-2hv_2/r^2$. Put $\Delta=u_1v_2-u_2v_1$, which is nonzero: Δ=0 would make the two direction vectors proportional and the chord lines through M identical. A line through $(x_1,y_1)$ and $(x_2,y_2)$ has equation $(y_2-y_1)x+(x_1-x_2)y=x_1y_2-x_2y_1$. Setting y=0 gives its horizontal intercept $(x_1y_2-x_2y_1)/(y_2-y_1)$ when that intercept exists. Applying this to AD and BC, their horizontal intercepts ξ and η satisfy $$\begin{gathered}\frac1\xi=\frac{dv_2-au_2}{ad\Delta},\\ \frac1\eta=\frac{cv_2-bu_2}{bc\Delta}.\end{gathered}$$ The intersections are finite by assumption, and nonzero because AD and BC do not pass through M. Adding now gives $$\frac1\xi+\frac1\eta=\frac{v_2(1/a+1/b)-u_2(1/c+1/d)}\Delta=0.$$ Thus ξ+η=0. Since M has horizontal coordinate zero and X,Y lie on the horizontal line, M is their midpoint.

Conclusion: The intersections X and Y have equal and opposite coordinates from M.

Review the idea: Circles and power of a point · Vietas formulas

Write S for the total. The subset sums are $2^{n}$ distinct integers in the S+1 available positions 0,1,…,S, so S≥$2^{n}$−1. If equality holds, all those positions occur. The value 1 must occur, forcing $a_{1}$=1. Suppose the first k entries are 1,2,…,$2^{k-1}$; they produce exactly the sums 0,…,$2^{k}$−1, each once, by induction. If the next entry were smaller than $2^{k}$, its singleton sum would duplicate one of those earlier sums. If it were larger than $2^{k}$, neither earlier entries, whose total is $2^{k}$−1, nor any later entry could form the missing sum $2^{k}$. Therefore the next entry equals $2^{k}$. This proves all entries have the claimed form. Conversely these powers of two work: when the next power $2^{k}$ is added, subsets omitting it have sums 0,…,$2^{k}$−1, and subsets containing it have sums $2^{k}$,…,$2^{k+1}$−1. The two ranges are disjoint and each sum is unique.

Conclusion: The minimum total is $2^{n}$−1, attained exactly by the successive powers of 2.

Review the idea: Counting · Strong induction

For a fraction strictly between the endpoints, let $u=bp-aq$ and $v=cq-dp$. Both are positive integers. The given identity bc−ad=1 gives $$q=(bc-ad)q=bv+du\ge b+d.$$ Equality holds only if u=v=1, since b,d are positive. Solving then gives $p=av+cu=a+c$; this identity follows by expanding the right side and using bc−ad=1. Thus the only possible numerator at denominator b+d is a+c. Conversely $b(a+c)-a(b+d)=bc-ad=1$ and $c(b+d)-d(a+c)=bc-ad=1$, so the mediant lies strictly between the endpoints. Any common divisor of a+c and b+d would divide the first displayed difference 1, proving that it is in lowest terms. This also covers negative numerators a or c.

Conclusion: The least denominator is b+d, uniquely attained by the reduced mediant.

Review the idea: Greatest common divisor · Exact arithmetic

Both α and β exceed 1, and β is irrational: if β were rational, α=β/(β−1) would be rational too. Each floor sequence is strictly increasing, since increasing n adds more than 1 before taking the floor. Fix an integer m≥0. The inequality floor(nα)≤m is equivalent to nα<m+1, or n<(m+1)/α. This upper endpoint is noninteger, so the number of positive n allowed is floor((m+1)/α). The analogous count for β is floor((m+1)/β). The two arguments have sum m+1 and neither is an integer; their fractional parts therefore sum to 1. Thus the total number of sequence terms at most m, with repetitions counted, is exactly m. It is zero at m=0. Subtracting the count for m−1 from the count for m shows that exactly one term equals each positive integer m. Hence there are neither omissions nor overlaps.

Conclusion: Together the two sequences contain each positive integer exactly once.

Review the idea: Floor and ceiling functions · Counting with bijections