MathDB

1

Part of 1997 Romania National Olympiad

Problems(4)

abcabc, d00d, radicals and integers - 1997 Romania NMO VII p1

Source:

8/13/2024
Let n1=abcabc‾n_1 = \overline{abcabc} and n2=d00d‾n_2= \overline{d00d} be numbers represented in the decimal system, with a≠0a\ne 0 and d≠0d \ne 0.
a) Prove that n1\sqrt{n_1} cannot be an integer. b) Find all positive integers n1n_1 and n2n_2 such that n1+n2\sqrt{n_1+n_2} is an integer number. c) From all the pairs (n1,n2)(n_1,n_2) such that n1n2\sqrt{n_1 n_2} is an integer find those for which n1n2\sqrt{n_1 n_2} has the greatest possible value
number theoryInteger
3 straight lines intersecting n circles, equal segments wanted and given

Source: 1997 Romania NMO IX p1

8/13/2024
Let C1,C2,...,CnC_1,C_2,..., C_n , (n≥3)(n\ge 3) be circles having a common point MM. Three straight lines passing through MM intersect again the circles in A1,A2,...,AnA_1, A_2,..., A_n ; B1,B2,...,BnB_1,B_2,..., B_n and X1,X2,...,XnX_1,X_2,..., X_n respectively. Prove that if A1A2=A2A3=...=An−1AnA_1A_2 =A_2A_3 =...=A_{n-1}A_n and B1B2=B2B3=...=Bn−1BnB_1B_2 =B_2B_3 =...=B_{n-1}B_n then X1X2=X2X3=...=Xn−1Xn.X_1X_2 =X_2X_3 =...=X_{n-1}X_n.
geometrycirclesequal segments
P(X) = X^{1997}-X^{1995} +X^2-3kX+3k+1 - 1997 Romania NMO VIΙI p1

Source:

8/13/2024
Let kk be an integer number and P(X)P(X) be the polynomial P(X)=X1997−X1995+X2−3kX+3k+1P(X) = X^{1997}-X^{1995} +X^2-3kX+3k+1 Prove that: a) the polynomial has no integer root; β) the numbers P(n)P(n) and P(n)+3P(n) + 3 are relatively prime, for every integer nn.
algebrapolynomial
find f(3,1997)

Source: Romania 1997

8/26/2005
function f:N⋆×N⋆→N⋆f:\mathbb{N}^{\star} \times \mathbb{N}^{\star} \rightarrow \mathbb{N}^{\star} (N⋆=N∪{0}\mathbb{N}^{\star}=\mathbb{N}\cup \{0\})with these conditon: 1- f(0,x)=x+1f(0,x)=x+1 2- f(x+1,0)=f(x,1)f(x+1,0)=f(x,1) 3- f(x+1,y+1)=f(x,f(x+1,y))f(x+1,y+1)=f(x,f(x+1,y))(romania 1997) find f(3,1997)f(3,1997)
functioninductionalgebrafunctional equationIMO 1981IMOromania