MathDB

1992 Hungary-Israel Binational

Part of Hungary-Israel Binational

Subcontests

(6)
6
1

rectangle has coordinates of vertices are Fibonacci numbers

We examine the following two sequences: The Fibonacci sequence: F0=0,F1=1,Fn=Fn−1+Fn−2F_{0}= 0, F_{1}= 1, F_{n}= F_{n-1}+F_{n-2 } for n≥2n \geq 2; The Lucas sequence: L0=2,L1=1,Ln=Ln−1+Ln−2L_{0}= 2, L_{1}= 1, L_{n}= L_{n-1}+L_{n-2} for n≥2n \geq 2. It is known that for all n≥0n \geq 0 Fn=αn−βn5,Ln=αn+βn,F_{n}=\frac{\alpha^{n}-\beta^{n}}{\sqrt{5}},L_{n}=\alpha^{n}+\beta^{n}, where α=1+52,β=1−52\alpha=\frac{1+\sqrt{5}}{2},\beta=\frac{1-\sqrt{5}}{2}. These formulae can be used without proof.
The coordinates of all vertices of a given rectangle are Fibonacci numbers. Suppose that the rectangle is not such that one of its vertices is on the xx-axis and another on the yy-axis. Prove that either the sides of the rectangle are parallel to the axes, or make an angle of 45∘45^{\circ} with the axes.
5
1

(...) can be written as a product of three Fibonacci numbers

We examine the following two sequences: The Fibonacci sequence: F0=0,F1=1,Fn=Fn−1+Fn−2F_{0}= 0, F_{1}= 1, F_{n}= F_{n-1}+F_{n-2 } for n≥2n \geq 2; The Lucas sequence: L0=2,L1=1,Ln=Ln−1+Ln−2L_{0}= 2, L_{1}= 1, L_{n}= L_{n-1}+L_{n-2} for n≥2n \geq 2. It is known that for all n≥0n \geq 0 Fn=αn−βn5,Ln=αn+βn,F_{n}=\frac{\alpha^{n}-\beta^{n}}{\sqrt{5}},L_{n}=\alpha^{n}+\beta^{n}, where α=1+52,β=1−52\alpha=\frac{1+\sqrt{5}}{2},\beta=\frac{1-\sqrt{5}}{2}. These formulae can be used without proof. Show that L2n+1+(−1)n+1(n≥1)L_{2n+1}+(-1)^{n+1}(n \geq 1) can be written as a product of three (not necessarily distinct) Fibonacci numbers.
3
2

On 100 strictly increasing sequences of positive integers

We are given 100100 strictly increasing sequences of positive integers: Ai=(a1(i),a2(i),...),i=1,2,...,100A_{i}= (a_{1}^{(i)}, a_{2}^{(i)},...), i = 1, 2,..., 100. For 1≤r,s≤1001 \leq r, s \leq 100 we define the following quantities: fr(u)=f_{r}(u)= the number of elements of ArA_{r} not exceeding nn; fr,s(u)=f_{r,s}(u) = the number of elements of Ar∩AsA_{r}\cap A_{s} not exceeding nn. Suppose that fr(n)≥12nf_{r}(n) \geq\frac{1}{2}n for all rr and nn. Prove that there exists a pair of indices (r,s)(r, s) with r≠sr \not = s such that fr,s(n)≥8n33f_{r,s}(n) \geq\frac{8n}{33} for at least five distinct n−sn-s with 1≤n<19920.1 \leq n < 19920.

r-Fibonacci number

We examine the following two sequences: The Fibonacci sequence: F0=0,F1=1,Fn=Fn−1+Fn−2F_{0}= 0, F_{1}= 1, F_{n}= F_{n-1}+F_{n-2 } for n≥2n \geq 2; The Lucas sequence: L0=2,L1=1,Ln=Ln−1+Ln−2L_{0}= 2, L_{1}= 1, L_{n}= L_{n-1}+L_{n-2} for n≥2n \geq 2. It is known that for all n≥0n \geq 0 Fn=αn−βn5,Ln=αn+βn,F_{n}=\frac{\alpha^{n}-\beta^{n}}{\sqrt{5}},L_{n}=\alpha^{n}+\beta^{n}, where α=1+52,β=1−52\alpha=\frac{1+\sqrt{5}}{2},\beta=\frac{1-\sqrt{5}}{2}. These formulae can be used without proof. We call a nonnegative integer rr-Fibonacci number if it is a sum of rr (not necessarily distinct) Fibonacci numbers. Show that there infinitely many positive integers that are not rr-Fibonacci numbers for any r,1≤r≤5.r, 1 \leq r\leq 5.
2
2

set of 1992 positive integers and sum of elements of subsets

A set SS consists of 19921992 positive integers among whose units digits all 1010 digits occur. Show that there is such a set SS having no nonempty subset S1S_{1} whose sum of elements is divisible by 20002000.

an equality on Fibonacci numbers

We examine the following two sequences: The Fibonacci sequence: F0=0,F1=1,Fn=Fn−1+Fn−2F_{0}= 0, F_{1}= 1, F_{n}= F_{n-1}+F_{n-2 } for n≥2n \geq 2; The Lucas sequence: L0=2,L1=1,Ln=Ln−1+Ln−2L_{0}= 2, L_{1}= 1, L_{n}= L_{n-1}+L_{n-2} for n≥2n \geq 2. It is known that for all n≥0n \geq 0 Fn=αn−βn5,Ln=αn+βn,F_{n}=\frac{\alpha^{n}-\beta^{n}}{\sqrt{5}},L_{n}=\alpha^{n}+\beta^{n}, where α=1+52,β=1−52\alpha=\frac{1+\sqrt{5}}{2},\beta=\frac{1-\sqrt{5}}{2}. These formulae can be used without proof. Prove that ∑k=1n[αkFk+12]=F2n+1  ∀n>1.\sum_{k=1}^{n}[\alpha^{k}F_{k}+\frac{1}{2}]=F_{2n+1}\; \forall n>1.
1
2

a+1/a-2

Prove that if cc is a positive number distinct from 11 and nn a positive integer, then n2≤cn+c−n−2c+c−1−2.n^{2}\leq \frac{c^{n}+c^{-n}-2}{c+c^{-1}-2}.

2^{j+1}|1+L_{2j} (Lucas sequence)

We examine the following two sequences: The Fibonacci sequence: F0=0,F1=1,Fn=Fn−1+Fn−2F_{0}= 0, F_{1}= 1, F_{n}= F_{n-1}+F_{n-2 } for n≥2n \geq 2; The Lucas sequence: L0=2,L1=1,Ln=Ln−1+Ln−2L_{0}= 2, L_{1}= 1, L_{n}= L_{n-1}+L_{n-2} for n≥2n \geq 2. It is known that for all n≥0n \geq 0 Fn=αn−βn5,Ln=αn+βn,F_{n}=\frac{\alpha^{n}-\beta^{n}}{\sqrt{5}},L_{n}=\alpha^{n}+\beta^{n}, where α=1+52,β=1−52\alpha=\frac{1+\sqrt{5}}{2},\beta=\frac{1-\sqrt{5}}{2}. These formulae can be used without proof. Prove that 1+L2j≡0(mod2j+1)1+L_{2^{j}}\equiv 0 \pmod{2^{j+1}} for j≥0j \geq 0.
4
2

Pentagon in the plane [lattice point in a pentagram]

We are given a convex pentagon ABCDEABCDE in the coordinate plane such that AA, BB, CC, DD, EE are lattice points. Let QQ denote the convex pentagon bounded by the five diagonals of the pentagon ABCDEABCDE (so that the vertices of QQ are the interior points of intersection of diagonals of the pentagon ABCDEABCDE). Prove that there exists a lattice point inside of QQ or on the boundary of QQ.

(...) is not a perfect square (Fibonacci and Lucas numbers)

We examine the following two sequences: The Fibonacci sequence: F0=0,F1=1,Fn=Fn−1+Fn−2F_{0}= 0, F_{1}= 1, F_{n}= F_{n-1}+F_{n-2 } for n≥2n \geq 2; The Lucas sequence: L0=2,L1=1,Ln=Ln−1+Ln−2L_{0}= 2, L_{1}= 1, L_{n}= L_{n-1}+L_{n-2} for n≥2n \geq 2. It is known that for all n≥0n \geq 0 Fn=αn−βn5,Ln=αn+βn,F_{n}=\frac{\alpha^{n}-\beta^{n}}{\sqrt{5}},L_{n}=\alpha^{n}+\beta^{n}, where α=1+52,β=1−52\alpha=\frac{1+\sqrt{5}}{2},\beta=\frac{1-\sqrt{5}}{2}. These formulae can be used without proof. Prove that Fn−1FnFn+1Ln−1LnLn+1(n≥2)F_{n-1}F_{n}F_{n+1}L_{n-1}L_{n}L_{n+1}(n \geq 2) is not a perfect square.