Problem 113

Suppose that, for some integer $n \geq 2$, we have $a_{n}=x$ and $a_{n-1}=y$. The equation $a_{n}+a_{n-1}=\frac{5}{2} \sqrt{a_{n} \cdot a_{n-1}}$ can be re-written as $x+y=\frac{5}{2} \sqrt{x y}$. Since $x>0$ and $y>0$, squaring both sides of the equation gives an equivalent equation which is $(x+y)^{2}=\frac{25}{4} x y$. Manipulating algebraically, we obtain the following equivalent equations: $(x+y)^{2}=\frac{25}{4} x y$, $4(x^{2}+2 x y+y^{2})=25 x y$, $4 x^{2}-17 x y+4 y^{2}=0$, $(4 x-y)(x-4 y)=0$. Therefore, the given relationship is equivalent to $4 x=y$ or $x=4 y$. Returning to the sequence notation, we now know that it is the case that $4 a_{n}=a_{n-1}$ (that is, $a_{n}=\frac{1}{4} a_{n-1}$ ) or $a_{n}=4 a_{n-1}$. Putting this another way, each term in the sequence can be obtained from the previous term either by multiplying by 4 or by dividing by 4. We are told that $a_{1}=4$ and $a_{11}=1024$. We note $\frac{a_{11}}{a_{1}}=\frac{1024}{4}=256=4^{4}$. We can think of moving along the sequence from $a_{1}$ to $a_{11}$ by making 10 "steps", each of which involves either multiplying by 4 or dividing by 4. If there are $m$ steps in which we multiply by 4 and $10-m$ steps in which we divide by 4, then $\frac{4^{m}}{4^{10-m}}=4^{4}$ which gives $4^{2 m-10}=4^{4}$ or $2 m-10=4$ and so $m=7$. In other words, the sequence involves 7 steps of multiplying by 4 and 3 steps of dividing by 4. These steps completely define the sequence. The number of possible sequences, $S$, equals the number of ways of arranging these 10 steps, which equals $\binom{10}{3}$. (If combinatorial notation is unfamiliar, we could systematically count the number of arrangements instead.) Therefore, $S=\frac{10 \cdot 9 \cdot 8}{3 \cdot 2 \cdot 1}=5 \cdot 3 \cdot 8=120$. The rightmost two digits of $S$ are 20.