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Baltic Way 1997 · Problem 3

Algebra

Let x1=1x_{1}=1 and xn+1=xn+⌊xnn⌋+2x_{n+1}=x_{n}+\left\lfloor\frac{x_{n}}{n}\right\rfloor+2 for n=1,2,3,…n=1,2,3, \ldots, where ⌊x⌋\lfloor x\rfloor denotes the largest integer not greater than xx. Determine x1997x_{1997}.

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Topics

Sequences and recurrences

Solutions

Solution

Solution: x1997=23913x_{1997}=23913.

Note that if xn=an+bx_{n}=a n+b with 0⩽b<n0 \leqslant b<n, then

xn+1=xn+a+2=a(n+1)+b+2.x_{n+1}=x_{n}+a+2=a(n+1)+b+2.

Hence if xN=ANx_{N}=A N for some positive integers AA and NN, then for i=0,1,…,Ni=0,1, \ldots, N we have xN+i=A(N+i)+2ix_{N+i}=A(N+i)+2 i, and x2N=(A+1)⋅2Nx_{2 N}=(A+1) \cdot 2 N.

Since for N=1N=1 the condition xN=ANx_{N}=A N holds with A=1A=1, then for N=2kN=2^{k} (where kk is any non-negative integer) it also holds with A=k+1A=k+1.

Now for N=210=1024N=2^{10}=1024 we have A=11A=11 and xN+i=A(N+i)+2ix_{N+i}=A(N+i)+2 i, which for i=973i=973 makes x1997=11⋅1997+2⋅973=23913x_{1997}=11 \cdot 1997+2 \cdot 973=23913.

Contest context

Results from Baltic Way 1997

11 teams

Mean score
2.5 / 5
Scores of 4 or 5
5 / 11
Estonia
1 / 5

Score distribution

03
12
20
31
43
52
All team scores
TeamScore
Poland4 / 5
Germany4 / 5
Estonia1 / 5
Sweden5 / 5
Denmark5 / 5
Latvia3 / 5
Finland0 / 5
Norway1 / 5
St. Petersburg0 / 5
Iceland0 / 5
Lithuania4 / 5