# Problem 9.

Let $A(n)$ denote the greatest odd divisor of the number $n$. For example, $A(21)=21$, $A(72)=9, A(64)=1$. Find the sum $A(111)+A(112)+\ldots+A(218)+A(219)$.
Answer: 12045

Solution. The largest odd divisors of any two of the given numbers cannot coincide, as numbers with the same largest odd divisors are either equal or differ by at least a factor of 2. Therefore, the largest odd divisors of the numbers $111, 112, \ldots, 218, 219$ are distinct. This means that the largest odd divisors of the numbers from $n+1$ to $2n$ are $n$ different odd numbers that do not exceed $2n$. Thus, these numbers are $1, 3, 5, \ldots, 2n-1$. If we add the number 220 to the set of numbers, the desired sum will be $1+3+5+\ldots+219-A(220)=\frac{1+219}{2} \cdot 110-55=110^2-55=12045$.

## B-2

Let $A(n)$ denote the largest odd divisor of the number $n$. For example, $A(21)=21$, $A(72)=9$, $A(64)=1$. Find the sum $A(113)+A(114)+\ldots+A(222)+A(223)$.

Answer: 12537

## B-3

Let $A(n)$ denote the largest odd divisor of the number $n$. For example, $A(21)=21$, $A(72)=9$, $A(64)=1$. Find the sum $A(115)+A(116)+\ldots+A(226)+A(227)$.

Answer: 12939

## B-4

Let $A(n)$ denote the largest odd divisor of the number $n$. For example, $A(21)=21$, $A(72)=9$, $A(64)=1$. Find the sum $A(117)+A(118)+\ldots+A(230)+A(231)$.

Answer: 13427