Problem 5

Given

n_0>1

, players A and B choose integers alternately. Knowing

n_{2k}

, A chooses

n_{2k+1}

with

n_{2k}\le n_{2k+1}\le n_{2k}^2

. Knowing

n_{2k+1}

, B chooses

n_{2k+2}

such that

n_{2k+1}/n_{2k+2}=p^r

for a prime

p

and integer

r\ge1

. A wins by choosing

1990

, and B wins by choosing

1

. Classify the initial values according to which player has a winning strategy or neither does.

Step 3 of 6: Handle values above 1990

Previous step Next step

n=1991=11\cdot181\mapsto1991,\qquad 11^r\cdot181+1\le n\le11^{r+1}\cdot181\mapsto11^{r+1}\cdot181

Detailed analysis

For $n=1991=11\cdot181$ , A repeats 1991. More generally, if $11^r\cdot181+1\le n\le11^{r+1}\cdot181$ , choose $11^{r+1}\cdot181$ . After B divides by a prime power, the result is strictly smaller than the current $n$ but remains at least $11$ ; repeated use reaches the earlier winning range.