AIMO 2015 Q4

Since $a+b=1024$ we can say that $\frac{a}{b}=\frac{a}{1024-a}$ leading to $\frac{a}{1024-a}=\frac{2n+1}{n^{2}}$ .

Cross multiplying gives us $an^{2}=2048n-2an+1024-a$ .

Bringing all the terms with $a$ to one side and factoring we get $a(n^{2}+2n+1)=1024(2n+1)$ .

Dividing through we get $a=\frac{1024(2n+1)}{(n+1)^{2}}$ .

Because $\frac{a}{b}$ is in lowest terms both $a$ and $b$ cannot be even. With the sum $a+b=1024$ we can see that both must be odd.

Because $a$ must be odd $1024$ must be completely divided by $(n+1)^{2}$ . Thus $32|n+1$ .

Let $k$ be the value such that $32k=n+1$ , thus $n=32k-1$ .

Substituting we get $a=\frac{1024(64k-1)}{1024k^{2}}=\frac{(64k-1)}{k^{2}}$ .

The only value for which this is an integer is $k=1$ .

Thus $n=31$ and $a=63$ .

$\frac{a}{b}=\frac{2}{n}+\frac{1}{n^2}=\frac{2n+1}{n^2}$

Now we can see that $a$ and $b$ are in the form of $2n+1$ and $n^2$ respectively.

$\begin{aligned} a+b & =1024 \\ n^2+2n+1 & =1024 \\ n^2+2n-1023 & =0 \end{aligned}$

Now we can solve the above quadratic equation and we'll get $n=-33,31$ . But $-33$ will be rejected as $n$ is a positive integer, so, $n=31$ .

Placing $n=31$ in $\frac{2n+1}{n^2}$ will give us the value of $a$ as $2n+1=\boxed{63}$ .

Rewrite the n terms into one fraction; $\frac {a}{b}=\frac {2n +1}{n^{2}}$ So we know that a is always odd, so for $a+b=1024= 2 ^{10}$ to hold true, b is odd. Thus in our case,

$n=2k +1 : k\in\mathbb {N_{0}}$

So we have

$a=4k + 3$ $b=4k^{2} +4k + 1$

Thus,

$a+b=4k +3+4k^{2} +4k + 1=2^{10}$ $k^{2} +2k + 1 =2^{8}$ $(k+1)^{2}=2^{8}$ So $k=15$ $n=31$ $a=63$

It is given that:

$\begin{aligned} \frac{2}{n} + \frac{1}{n^2} & = \frac{a}{b} \quad \quad \small \color{#3D99F6}{\text{and that } a+b = 1024} \\ \frac{2n + 1}{n^2} & = \frac{1024-b}{b} \\ \frac{2n + 1}{n^2} & = \frac{1024}{b} - 1 \\ \frac{2n + 1}{n^2} + 1 & = \frac{1024}{b} \\ \frac{2n + 1 + n^2}{n^2} & = \frac{1024}{b} \\ \left(\frac{n + 1}{n}\right)^2 & = \left(\frac{32}{\sqrt{b}}\right)^2 \end{aligned}$

$\Rightarrow n = 31 \text{ and } b = 31^2 = 961 \quad \Rightarrow a = 1024 - 961 = \boxed{63}$

First, add together $\frac{2}{n}$ and $\frac{1}{n^{2}}$ to get a single fraction. Given that $\frac{a}{b}$ is the reduced form of $\frac{2}{n}$ + $\frac{1}{n^2}$ , $\frac{2}{n}$ + $\frac{1}{n^2}$ = $\frac{a}{b}$ .

$\frac{2}{n}$ + $\frac{1}{n^2}$ = $\frac{a}{b}$

$\frac{2n+1}{n^{2}}$ = $\frac{a}{b}$

Substituting the numerators and denominators into a and b in $a+b=1024$ , we get $n^{2} + 2n +1 =1024$ . This is a quadratic that we can solve, and factoring this equation, we now have $(n-31)(n+33)=1024$ .

Finally, we solve for the n values, and we get $n=31$ and $n=-33$ . But which n value is the one we want? Well, if we plug the two values into $\frac{2n+1}{n^{2}}$ = $\frac{a}{b}$ , only $n=31$ will yield a positive a. value (stated in the problem). Thus, $a=2(31)+1= \boxed{63}$ .