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BdMO National 2013: Junior 9

Posted: Fri Jan 10, 2014 1:29 am
by BdMO
The ratio of GCD and LCM of two integers is $1: 36$ and sum of the integers is $5460$. What is the difference between these two integers?

Re: BdMO National 2013: Junior 9

Posted: Sun Jan 12, 2014 12:38 am
by Tahmid
let two numbers are a and b.
and let $\left ( a,b \right )=x \Leftrightarrow a=xa_{1}$ and $b=xb_{1}$ where $\left ( a_{1},b_{1} \right )=1$
so,$\left [ a,b \right ]=xa_{1}b_{1}$

but given that, \[ \frac{(a,b)}{[a,b]}=\frac{1}{36} \]
\[\frac{x}{xa_{1}b_{1}}=\frac{1}{36}\Leftrightarrow \frac{1}{a_{1}b_{1}}=\frac{1}{36}\Leftrightarrow a_{1}b_{1}=36\]

so now we get,$a_{1}b_{1}=36$ and $(a_{1},b_{1})=1$
the pair which satisfy these two property of $a_{1}$ and $b_{1}$ is $(4,9)$; that means $a_{1}=4$ and $b_{1}=9$

$\therefore a=4x$ and $b=9x$
given that $a+b=5460$
so, $a+b=5460\Leftrightarrow 4x+9x=5460\Leftrightarrow 13x=5460\Leftrightarrow x=420$

finally $x=420;a_{1}=4;b_{1}=9$

so $a=420*4=1680$ and $b=420*9=3780$
and difference $=3780-1680=2100$......(ans:).... :)

Re: BdMO National 2013: Junior 9

Posted: Sun Jan 12, 2014 2:07 am
by Labib
Another possible pair satisfying the conditions for $(a_1, b_1)$ is $(1,36)$.
Although it does not lead to any significant discovery, ignoring it might mean you losing out on some easy points for the problem.

Re: BdMO National 2013: Junior 9

Posted: Sun Jan 12, 2014 9:46 am
by Tahmid
i noticed that. but for pair (1,36), our next equation comes 37x=5460. but 37 does not divide 5460. then x becomes fraction. but x must be a integer. so (1,36) is not possible. i forget to include these steps in my solution :?

Re: BdMO National 2013: Junior 9

Posted: Mon Jan 13, 2014 7:31 pm
by Labib
Yes, that pair does not lead us to a solution. But it is a good practice to include these small details in your solution.
Otherwise you will lose 1 or 2 points in BdMO/BdMC/IMO. It was just a heads up. :)
I request, if it's still possible, please hide the solutions. It helps others.

Re: BdMO National 2013: Junior 9

Posted: Fri May 07, 2021 9:34 pm
by Swapnil Barua
In 2 numbers GCD their common factor remains multiple of those numbers and in LCM it remains with the common factors and other factors remains multiple. So in the ratio of GCD and LCM common factors cancel out and remains the others factors without those common factors.
Now, 36=2×2×3×3 these 4 factors can be of similar numbers [case 1] or 2×2 factor is of 1 number and 3×3 is of another.[case 2]
First take these 4 factors are of similar numbers, If 1 number is 36x another is x
According to the question
36x+x= 5460
or, 37x= 5460
or,x= 147.57 [ Not an integer]
Now take the 2nd case,
If 1 number is 4x another is 6x
According to the question
4x+6x=5460
or, 10x= 5460
or, x= 546
Difference=6x-4x=2x= 2× 546= 1092

Re: BdMO National 2013: Junior 9

Posted: Wed Jul 14, 2021 6:45 pm
by wick09788
Swapnil Barua wrote:
Fri May 07, 2021 9:34 pm
Fm Whatsapp apk
In 2 numbers GCD their common factor remains multiple of those numbers and in LCM it remains with the common factors and other factors remains multiple. So in the ratio of GCD and LCM common factors cancel out and remains the others factors without those common factors.
Now, 36=2×2×3×3 these 4 factors can be of similar numbers [case 1] or 2×2 factor is of 1 number and 3×3 is of another.[case 2]
First take these 4 factors are of similar numbers, If 1 number is 36x another is x
According to the question FmWhatsapp apk
36x+x= 5460
or, 37x= 5460
or,x= 147.57 [ Not an integer]
Now take the 2nd case,
If 1 number is 4x another is 6x
According to the question
4x+6x=5460
or, 10x= 5460
or, x= 546
Difference=6x-4x=2x= 2× 546= 1092

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Re: BdMO National 2013: Junior 9

Posted: Mon Jan 24, 2022 4:55 pm
by im2033thomas
Thanks for this solution, I was finding this on internet from long ago, And now I found this correct one. Thank You..