BdMO 2010 higher secondary Q. 6
$a$ 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: $201*402*10$
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: $201*402*10$
"Go down deep enough into anything and you will find mathematics." ~Dean Schlicter
Re: BdMO 2010 higher secondary Q. 6
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
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
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Re: BdMO 2010 higher secondary Q. 6
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 ????
২০১ টা সংখ্যা পাওয়া যায় এটা বুঝছি কিন্তু ১০ আর ৪০২ কেন আসল ????
হার জিত চিরদিন থাকবেই
তবুও এগিয়ে যেতে হবে.........
বাধা-বিঘ্ন না পেরিয়ে
বড় হয়েছে কে কবে.........
তবুও এগিয়ে যেতে হবে.........
বাধা-বিঘ্ন না পেরিয়ে
বড় হয়েছে কে কবে.........
Re: BdMO 2010 higher secondary Q. 6
I'm confoused. Apu, I think you have counted (a^2,b^2). But we have to count (a,b). @Urmi apu
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Re: BdMO 2010 higher secondary Q. 6
@Nadim,
$a \ne b$ if this condition is maintained, then the ans will be $2.(804)^2 + 401.400$.
$a \ne b$ if this condition is maintained, then the ans will be $2.(804)^2 + 401.400$.
Bristy Sikder
Re: BdMO 2010 higher secondary Q. 6
Why did you considered $a,b$ of the forms $5n+2,5n+3$ ? Because $(5n+2)^{2}+(5n+3)^{2}\equiv 3(mod 5)$.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 $:)
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Re: BdMO 2010 higher secondary Q. 6
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, $S\equiv 0(mod 5)$.SANZEED wrote:Why did you considered $a,b$ of the forms $5n+2,5n+3$ ? Because $(5n+2)^{2}+(5n+3)^{2}\equiv 3(mod 5)$.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 $:)
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 $4.402^2+401.400$