I am trying these two problem for many days . Can anyone help me? :-
Problem no.1 : Prove that for all integer $x$ and $y$ there is no solution for $15x^2 - 7y^2 = 9$
Problem no.2 : $X$ is a two digit positive integer and $Y$ is a three digit positive integer. If $X$ is increased by $Y%$ and $Y$ is decreased by $X%$, they results same. How many solutions of $(X, Y)$ are there for this case?(from the BdMO '16 divisional)
Need help to solve number theory problems
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- Thamim Zahin
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Re: Need help to solve number theory problems
In problem (1) take mod $5$. And remember that $y^2 \equiv 0,1,4$ $(mod $ $ 5)$.
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- Thanic Nur Samin
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Re: Need help to solve number theory problems
For problem 2:
Please check your statement. Missing a $ \textbf{%} $ would certainly not help people who want to answer your question.
From the question, it follows that,
\[X\left( 1+\dfrac{Y}{100}\right) = Y\left( 1-\dfrac{X}{100} \right) \]
\[XY=50Y-50X\]
\[(50-X)(50+Y)=2500\]
Note that $50-X$ is positive, as the other two are positive. Let $A=50-X$ and $B=50+Y$. $X\ge 10 \Rightarrow A\le 40$ and $100\le Y\le 999 \Rightarrow 150\le B\le 1049$.
By considering all the cases when the product of two positive integers is equal to $2500$ it is straightforward to deduce that the number of solutions is $\boxed{3}$, Namely $(X,Y)=(40,200),(45,450),(46,575)$
Please check your statement. Missing a $ \textbf{%} $ would certainly not help people who want to answer your question.
From the question, it follows that,
\[X\left( 1+\dfrac{Y}{100}\right) = Y\left( 1-\dfrac{X}{100} \right) \]
\[XY=50Y-50X\]
\[(50-X)(50+Y)=2500\]
Note that $50-X$ is positive, as the other two are positive. Let $A=50-X$ and $B=50+Y$. $X\ge 10 \Rightarrow A\le 40$ and $100\le Y\le 999 \Rightarrow 150\le B\le 1049$.
By considering all the cases when the product of two positive integers is equal to $2500$ it is straightforward to deduce that the number of solutions is $\boxed{3}$, Namely $(X,Y)=(40,200),(45,450),(46,575)$
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Re: Need help to solve number theory problems
How did you get that?
\[X\left(1+\frac{y}{100}\right)=Y\left(1-\frac{X}{100}\right)\]
\[X\left(1+\frac{y}{100}\right)=Y\left(1-\frac{X}{100}\right)\]
Last edited by Phlembac Adib Hasan on Fri Dec 09, 2016 3:22 pm, edited 1 time in total.
Reason: Latexed
Reason: Latexed
- Thanic Nur Samin
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Re: Need help to solve number theory problems
Umm, $X$ is increased by $Y\%$ and $Y$ is decreased by $X\%$. The equation naturally follows.
Hammer with tact.
Because destroying everything mindlessly isn't cool enough.
Because destroying everything mindlessly isn't cool enough.
Re: Need help to solve number theory problems
In problem 2;answer is 2
(40,200),(45,450)
(40,200),(45,450)