Rational

25 results

Croatian Mathematical Olympiad 2019 Problem M-1

Odredi sve funkcije f:Q+Q+f: \mathbb{Q}^+ \to \mathbb{Q}^+ takve da vrijedi f(x2(f(y))2)=(f(x))2f(y),za sve x,yQ+.f(x^2 (f(y))^2) = (f(x))^2 f(y), \quad \text{za sve } x, y \in \mathbb{Q}^+.

(Q+\mathbb{Q}^+ je oznaka za skup svih pozitivnih racionalnih brojeva.)

Croatian Mathematical Olympiad 2022 Problem M-1

Neka je Q0+\mathbb{Q}_0^+ skup svih nenegativnih racionalnih brojeva.

Odredi sve funkcije f:Q0+Q0+f: \mathbb{Q}_0^+ \to \mathbb{Q}_0^+ takve da za sve nenegativne racionalne brojeve xx, yy vrijedi

yf(x+y)+(y1)f(xy)=f(y2)f(x+1).yf(x + y) + (y - 1)f(xy) = f(y^2)f(x + 1).

Croatian Mathematical Olympiad 2023 Problem 1-4

Za pozitivan racionalan broj qq kažemo da je sjajan ako za svaki pozitivan racionalan broj xx postoje cijeli broj n0n \geqslant 0 i cijeli brojevi a0,,ana_0, \ldots, a_n takvi da je

x=qa0(q+1)a1(q+n)an.x = q^{a_0} \cdot (q + 1)^{a_1} \cdot \ldots \cdot (q + n)^{a_n}.

Odredi sve sjajne brojeve.

International Mathematical Olympiad 2008 Problem 2

(a) Prove that x2(x1)2+y2(y1)2+z2(z1)21\frac{x^2}{(x-1)^2} + \frac{y^2}{(y-1)^2} + \frac{z^2}{(z-1)^2} \geq 1 for all real numbers x,y,zx, y, z, each different from 1, and satisfying xyz=1xyz = 1.

(b) Prove that equality holds above for infinitely many triples of rational numbers x,y,zx, y, z, each different from 1, and satisfying xyz=1xyz = 1.

International Mathematical Olympiad 2013 Problem 5

Let Q>0\mathbb{Q}_{>0} be the set of positive rational numbers. Let f:Q>0Rf: \mathbb{Q}_{>0} \to \mathbb{R} be a function satisfying the following three conditions:

(i) for all x,yQ>0x, y \in \mathbb{Q}_{>0}, we have f(x)f(y)f(xy)f(x)f(y) \geq f(xy);

(ii) for all x,yQ>0x, y \in \mathbb{Q}_{>0}, we have f(x+y)f(x)+f(y)f(x + y) \geq f(x) + f(y);

(iii) there exists a rational number a>1a > 1 such that f(a)=af(a) = a.

Prove that f(x)=xf(x) = x for all xQ>0x \in \mathbb{Q}_{>0}.

International Mathematical Olympiad 2024 Problem 6

Let Q\mathbb{Q} be the set of rational numbers. A function f ⁣:QQf\colon \mathbb{Q} \to \mathbb{Q} is called aquaesulian if the following property holds: for every x,yQx, y \in \mathbb{Q}, f(x+f(y))=f(x)+yorf(f(x)+y)=x+f(y).f(x + f(y)) = f(x) + y \quad \text{or} \quad f(f(x) + y) = x + f(y).

Show that there exists an integer cc such that for any aquaesulian function ff there are at most cc different rational numbers of the form f(r)+f(r)f(r) + f(-r) for some rational number rr, and find the smallest possible value of cc.

Middle European Mathematical Olympiad 2018 Problem I-1

Let Q+\mathbb{Q}^+ denote the set of all positive rational numbers and let αQ+\alpha \in \mathbb{Q}^+. Determine all functions f ⁣:Q+(α,+)f\colon \mathbb{Q}^{+}\to (\alpha , + \infty) satisfying

f(x+yα)=f(x)+f(y)α,for allx,yQ+.f \left(\frac {x + y}{\alpha}\right) = \frac {f (x) + f (y)}{\alpha}, \quad \text {for all} \, x, y \in \mathbb {Q} ^ {+}.

Middle European Mathematical Olympiad 2018 Problem T-2

Let P(x)P(x) be a polynomial of degree n2n \geq 2 with rational coefficients such that P(x)P(x) has nn pairwise different real roots forming an arithmetic progression. Prove that among the roots of P(x)P(x) there are two that are also the roots of some polynomial of degree 22 with rational coefficients.

Grade 9 2005 Problem 3

Koju najveću vrijednost može poprimiti izraz 1k+1m+1n,\frac{1}{k} + \frac{1}{m} + \frac{1}{n}, ako su kk, mm, nn prirodni brojevi takvi da je 1k+1m+1n<1\dfrac{1}{k} + \dfrac{1}{m} + \dfrac{1}{n} < 1.

Grade 9 2020 Problem 1

U ovisnosti o realnom parametru mm odredi za koje realne brojeve xx vrijedi

xmx2+x2(1mx)+m.\frac{x - m}{x^2} + x \geqslant 2 \left(1 - \frac{m}{x}\right) + m.

Grade 9 2020 Problem 2

Odredi sve uređene trojke (a,b,c)(a, b, c) prirodnih brojeva za koje vrijedi abca \leqslant b \leqslant c i

37=1a+1ab+1abc.\frac{3}{7} = \frac{1}{a} + \frac{1}{ab} + \frac{1}{abc}.

Grade 10 2026 Problem 7

Neka su xx i yy racionalni brojevi takvi da su x+yx + y i x2+y2x^2 + y^2 cijeli brojevi. Jesu li nužno xx i yy cijeli brojevi?

Grade 11 1996 Problem 4

Neka su α\alpha i β\beta pozitivni iracionalni brojevi takvi da je 1α+1β=1\frac{1}{\alpha} + \frac{1}{\beta} = 1, te A={[nα]nN}A = \{[n\alpha] | n \in \mathbf{N}\} i B={[nβ]nN}B = \{[n\beta] | n \in \mathbf{N}\}. Dokažite da je tada AB=NA \cup B = \mathbf{N} i AB=A \cap B = \emptyset.

Naputak: Možete dokazati ekvivalentnu tvrdnju: Za funkciju π:NN\pi : \mathbf{N} \longrightarrow \mathbf{N} defini-ranu sa π(m)=Card{kkN,km,kA}+Card{kkN,km,kB}\pi(m) = \operatorname{Card}\{k | k \in \mathbf{N}, k \leq m, k \in A\} + \operatorname{Card}\{k | k \in \mathbf{N}, k \leq m, k \in B\} vrijedi π(m)=m\pi(m) = m, mN\forall m \in \mathbf{N}.

([x][x] je oznaka za najveći cijeli broj koji nije veći od xx.)

Grade 11 2023 Problem 4

Odredi sve racionalne brojeve xx za koje vrijedi xx(xx)=254.x \cdot \lfloor x \rfloor \cdot (x - \lfloor x \rfloor) = 254.

Za racionalni broj tt, t\lfloor t \rfloor najveći je cijeli broj koji nije veći od tt. Na primjer, 3.14=3\lfloor 3.14 \rfloor = 3, 3.14=4\lfloor -3.14 \rfloor = -4.

Grade 12 1996 Problem 4

Neka su α\alpha i β\beta pozitivni iracionalni brojevi i 1α+1β=1\frac{1}{\alpha} + \frac{1}{\beta} = 1, te A={[nα]nN}A = \{[n\alpha]|n \in \mathbb{N}\} i B={[nβ]nN}B = \{[n\beta]|n \in \mathbb{N}\}. Dokažite da je tada AB=NA \cup B = \mathbb{N} i AB=A \cap B = \emptyset.

Naputak: Možete dokazati ekvivalentnu tvrdnju: Za funkciju π:NN\pi : \mathbb{N} \longrightarrow \mathbb{N} defini-ranu sa π(m)=Card{kkN,km,kA}+Card{kkN,km,kB}\pi(m) = \operatorname{Card}\{k \mid k \in \mathbb{N}, k \leq m, k \in A\} + \operatorname{Card}\{k \mid k \in \mathbb{N}, k \leq m, k \in B\} vrijedi π(m)=m\pi(m) = m, mN\forall m \in \mathbb{N}.

Grade 12 2011 Problem 3

Na koliko načina se broj 20112010\dfrac{2011}{2010} može prikazati kao umnožak dvaju razlomaka oblika n+1n\dfrac{n + 1}{n}, gdje je nn prirodan broj? Poredak faktora nije bitan.

Grade 12 2026 Problem 2

Neka je zz kompleksan broj takav da je broj 6z2+5z+63z2+10z+3\frac{6z^2 + 5z + 6}{3z^2 + 10z + 3} realan. Dokaži da je zz realan broj ili da vrijedi z=1|z| = 1.