## BdMO National 2013: Junior 9

BdMO
Posts: 134
Joined: Tue Jan 18, 2011 1:31 pm

### BdMO National 2013: Junior 9

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?

Tahmid
Posts: 110
Joined: Wed Mar 20, 2013 10:50 pm

### Re: BdMO National 2013: Junior 9

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:).... Labib
Posts: 411
Joined: Thu Dec 09, 2010 10:58 pm

### Re: BdMO National 2013: Junior 9

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.
Please Install $L^AT_EX$ fonts in your PC for better looking equations,
Learn how to write equations, and don't forget to read Forum Guide and Rules.

"When you have eliminated the impossible, whatever remains, however improbable, must be the truth." - Sherlock Holmes

Tahmid
Posts: 110
Joined: Wed Mar 20, 2013 10:50 pm

### Re: BdMO National 2013: Junior 9

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 Labib
Posts: 411
Joined: Thu Dec 09, 2010 10:58 pm

### Re: BdMO National 2013: Junior 9

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.
Please Install $L^AT_EX$ fonts in your PC for better looking equations,
Learn how to write equations, and don't forget to read Forum Guide and Rules.

"When you have eliminated the impossible, whatever remains, however improbable, must be the truth." - Sherlock Holmes

Swapnil Barua
Posts: 14
Joined: Tue Apr 13, 2021 2:55 pm

### Re: BdMO National 2013: Junior 9

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

wick09788
Posts: 1
Joined: Wed Jul 14, 2021 6:41 pm

### Re: BdMO National 2013: Junior 9

Swapnil Barua wrote:
Fri May 07, 2021 9:34 pm
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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

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