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$a$ and $b$ are two positive integers both less than $2010$; $a$ Find the number of ordered
pairs $(a, b)$ such that $a^{2} + b^{2}$ is divisible by $5$. Find $a + b$ so that $a^{2} + b^{2}$ is maximum.

Is the solutions correct? :S

The last digit $d$ of $a^{2}$ or $b^{2}$ is from the set $S$ = $\{ d \in d: a^{2} \equiv d (mod 10)\}$ or $S$ = $\{1, 4, 9, 6, 5, 6, 9, 4, 1, 0\}$. There are $201$ digits ending with each of these numbers.
Now we are to find pairs $(a, b)$ such that their last digits are $(1, 4), (1, 9), (4, 6), (9, 1), (6, 4), (5, 5), (5, 0),(9, 1), (4, 6), (1, 4), (1, 9), (0, 5), (0, 0)$. Note: I am dealing only with the last digits because that determines it's divisibility by 5.

So, for every digit ending with 1 number from $S$, we get 2 possible pairs.
As $(a, b)$ is ordered, $(16, 8)$ and $(8, 16)$ are different. We will count by fixing a number for $a$ and counting possible $b$'s. Hence, as we are fixing one distinct number for $a$ each time, no double counting occurs.

Answer: $201*402*10$

If the condition $a \ne b$ is not imposed, the result is correct.
However, dealing with quadratic residue of $5$ gives a much direct access to the problem. What Urmi did is just a "broken into pieces" method of the easier one.

Nice solution

সব বুঝলাম কিন্তু শেষে কেন $201*402*10$ লেখা হল বুঝলাম না।
২০১ টা সংখ্যা পাওয়া যায় এটা বুঝছি কিন্তু ১০ আর ৪০২ কেন আসল ????

I'm confoused. Apu, I think you have counted (a^2,b^2). But we have to count (a,b). @Urmi apu

@Nadim,
$a \ne b$ if this condition is maintained, then the ans will be $2.(804)^2 + 401.400$.

Nadim Ul Abrar wrote:$a^2\equiv0^2,1^2,2^2,3^2,4^2\equiv0,1,-1,-1,1 (mod5)$

There are $804$ numbers of form $5n+2$ and $5n+3$ less then $2010$.
There are $804$ numbers of form $5n+1$ and $5n+4$ less then $2010$.
There are $401$ numbers of form $5n$ less then $2010$.

So that number of pair $(a,b)$ be $2.804^2+401^2$.

And (a^2+b^2)_{max} can be found for the pair $(2009,2008),(2008,2009)$ , So that required value of $a+b$ be $4017 $:)

Why did you considered $a,b$ of the forms $5n+2,5n+3$ ? Because $(5n+2)^{2}+(5n+3)^{2}\equiv 3(mod 5)$.

SANZEED wrote:

Nadim Ul Abrar wrote:$a^2\equiv0^2,1^2,2^2,3^2,4^2\equiv0,1,-1,-1,1 (mod5)$

There are $804$ numbers of form $5n+2$ and $5n+3$ less then $2010$.
There are $804$ numbers of form $5n+1$ and $5n+4$ less then $2010$.
There are $401$ numbers of form $5n$ less then $2010$.

So that number of pair $(a,b)$ be $2.804^2+401^2$.

And $(a^2+b^2)_{max}$ can be found for the pair $(2009,2008),(2008,2009)$ , So that required value of $a+b$ be $4017 $:)

Why did you considered $a,b$ of the forms $5n+2,5n+3$ ? Because $(5n+2)^{2}+(5n+3)^{2}\equiv 3(mod 5)$.

The pairs Nadim Bhai should be considering are $5n+1$ and $5n+2$, $5n+1$ and $5n+3$, $5n+2$ and $5n+4$, $5n+3$ and $5n+4$. No matter which pair you take the sum of the squares, $S\equiv 0(mod 5)$.

There are $402$ numbers of form $5n+1$, $402$ numbers of form $5n+2$, $402$ numbers of form $5n+3$ and $402$ numbers of form $5n+4$ less then $2010$.
There are $401$ numbers of form $5n$ less then $2010$.

As $a \ne b$ total ordered pairs should be $4.402^2+401.400$