Primary Divisonal 2012/4
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Please don't post problems (by starting a topic) in the "Primary: Solved" forum. This forum is only for showcasing the problems for the convenience of the users. You can post the problems in the main Divisional Math Olympiad forum. Later we shall move that topic with proper formatting, and post in the resource section.
- Phlembac Adib Hasan
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The LCM of two numbers is $7$ times of their GCD. If the sum of the numbers is $56$, find their GCD.
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Re: Primary Divisonal 2012/4
The GCD is $7$.
Soppose,the numbers are $a$ and $b$ and suppose the GCD of the numbers are $x$.So,$a=xa_{1}$ and $b=xb_{1}$ for some integers $a_{1}$ and $b_{1}$ so that the GCD of $a_{1}$ and $b_{1}$ is $1$
$\therefore$ $7x=xa_{1}b_{1}$.Or,$a_{1}b_{1}=7$.So,one of $a_{1}$ and $b_{1}$ is $1$ and other is $7$.
Now,$xa_{1}+xb_{1}=56$
Or,$x(a_{1}+b_{1})=56$
Or,$8x=56$
Or,$x=7$
Soppose,the numbers are $a$ and $b$ and suppose the GCD of the numbers are $x$.So,$a=xa_{1}$ and $b=xb_{1}$ for some integers $a_{1}$ and $b_{1}$ so that the GCD of $a_{1}$ and $b_{1}$ is $1$
$\therefore$ $7x=xa_{1}b_{1}$.Or,$a_{1}b_{1}=7$.So,one of $a_{1}$ and $b_{1}$ is $1$ and other is $7$.
Now,$xa_{1}+xb_{1}=56$
Or,$x(a_{1}+b_{1})=56$
Or,$8x=56$
Or,$x=7$
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